Integration By Parts
Summary
TLDRThis educational video script covers the technique of integration by parts to solve various integral problems. It explains the process step by step, starting with the integral of x times e^x and progressing to more complex examples such as the integral of x^2 ln(x) and combinations of exponential and trigonometric functions. The script also tackles integrals involving logarithmic functions and provides a method to solve definite integrals, exemplified by e^(3x)cos(4x). Each example builds on the previous, reinforcing the concept of integration by parts for students to master calculus integration.
Takeaways
- 📚 Integration by parts is a fundamental technique for solving integrals of products of functions, expressed as ∫udv = uv - ∫vdu.
- 🔍 Choosing 'u' and 'dv' wisely is crucial; typically, 'u' is chosen to simplify when differentiated, and 'dv' is chosen to be easily integrable.
- 📈 The integral of x * e^x is found by setting x as 'u' and e^x dx as 'dv', resulting in x * e^x - e^x + C.
- 🌀 For integrals involving trigonometric functions, such as x * sin(x), setting x as 'u' and sin(x) dx as 'dv' leads to -x * cos(x) + sin(x) + C.
- 📘 When integrating a product of a polynomial and a logarithmic function, like x^2 * ln(x), it's often necessary to apply integration by parts multiple times.
- ⚡ The integral of ln(x) is x * ln(x) - x + C, which is derived using integration by parts and recognizing the integral of 1 dx as x.
- 🔗 The integral of x^2 * sin(x) involves using integration by parts twice and results in a complex expression involving both sine and cosine functions.
- 🌟 The integral of x^2 * e^x requires applying integration by parts twice, leading to a solution involving powers of e^x.
- 📉 The integral of ln(x) squared involves rewriting the expression and applying integration by parts, resulting in x * ln(x) squared - 2x * ln(x) + 2x + C.
- 📊 The integral of ln(x) to the seventh power is simplified by recognizing it as 7 * ln(x), leading to 7 * (x * ln(x) - x) + C.
- 🌌 The integral of e^x * sin(x) is solved by recognizing the pattern in the integral and using a clever manipulation to find the solution as (1/2) * e^x * (sin(x) - cos(x)) + C.
Q & A
What is the integral of x times e^x?
-The integral of x times e^x is found using integration by parts, setting u=x and dv=e^x dx, which gives du=dx and v=e^x. The result is x * e^x - integral of e^x dx, which simplifies to x * e^x - e^x + C, where C is the constant of integration.
How do you integrate x times sine x dx?
-For x times sine x, set u=x and dv=sine x dx, giving du=dx and v=-cosine x. Using integration by parts, the integral becomes -x * cosine x + integral of cosine x dx, which simplifies to -x * cosine x + sine x + C.
What is the antiderivative of x squared times ln x?
-To find the antiderivative of x squared times ln x, set dv=x squared dx and u=ln x, giving v=x cubed / 3 and du=1/x dx. The integral by parts results in (1/3) * x cubed * ln x - (1/3) * integral of x squared dx, which simplifies to (1/3) * x cubed * ln x - (1/9) * x cubed + C.
How can you find the integral of the natural log of x using integration by parts?
-For the natural log of x, set u=ln x and dv=dx, giving du=1/x dx and v=x. The integral by parts formula yields x * ln x - integral of x * (1/x) dx, which simplifies to x * ln x - x + C.
What is the indefinite integral of x squared times sine x dx?
-For x squared times sine x, you need to use integration by parts twice. Initially, set u=x squared and dv=sine x dx, giving du=2x dx and v=-cosine x. The process involves further integration by parts on the resulting integral of 2x * cosine x dx, leading to a final answer involving x squared * cosine x, 2x * sine x, and 2 * cosine x, plus the constant C.
How do you integrate x squared times e^x dx?
-For x squared times e^x, use integration by parts twice. Start by setting u=x squared and dv=e^x dx, giving du=2x dx and v=e^x. The process involves another round of integration by parts with u=2x and dv=e^x dx, leading to a final expression involving e^x terms and a constant C.
What is the integral of ln x squared dx?
-To integrate ln x squared, set u=ln x squared and dv=dx, giving du=2 * ln x * (1/x) dx and v=x. The integral by parts results in x * ln x squared - 2 integral of ln x * (1/x) dx. Solving this requires recognizing the integral of ln x and simplifying to get the final answer involving x * ln x squared, -2x * ln x, and 2x, plus C.
How can you rewrite the integral of ln x to the seventh power dx?
-The integral of ln x to the seventh power can be rewritten using the property of logs to move the exponent to the front, resulting in 7 * integral of ln x dx. Since the integral of ln x is x * ln x - x, the final answer is 7 * (x * ln x - x) + C.
What is the indefinite integral of e^x times sine x dx?
-For e^x times sine x, set u=sine x and dv=e^x dx, giving du=cosine x dx and v=e^x. Using integration by parts, the process involves recognizing the similarity of the resulting terms to the original integral and solving for the integral, leading to a final answer involving e^x * (sine x - cosine x) / 2 plus C.
How do you find the antiderivative of ln x squared divided by x dx?
-To find the antiderivative of ln x squared / x, rewrite it as ln x squared * 1/x dx and use integration by parts with u=ln x squared and dv=1/x dx, giving du=2 * ln x * (1/x) dx and v=ln x. The process involves recognizing the similarity of terms and solving for the integral, resulting in ln x cubed / 3 plus C.
What is the indefinite integral of e^(3x) times cosine 4x dx?
-For e^(3x) times cosine 4x, set u=cosine 4x and dv=e^(3x) dx, giving du=-sine 4x * 4 dx and v=(1/3)e^(3x). Using integration by parts, the process involves moving constants to the front and recognizing the similarity of terms, leading to a final answer involving e^(3x) * (cosine 4x / 3 + 4/3 * sine 4x) plus C.
Outlines
📚 Introduction to Integration by Parts
This paragraph introduces the concept of integration by parts, a technique used to find the integral of a product of two functions. It explains the formula for integration by parts, \( \int u \, dv = uv - \int v \, du \), and illustrates the process by setting up the functions 'u' and 'dv' for the integral of \( x \cdot e^x \). The explanation includes choosing 'u' and 'dv', finding 'du' and 'v', and applying the formula to arrive at the integral of \( x \cdot e^x \). It also briefly touches on the integral of \( x \cdot \sin(x) \) and \( x^2 \ln(x) \) as further examples.
🔍 Detailed Walkthrough of Integration by Parts
The second paragraph delves deeper into the application of integration by parts with the integral of \( x \cdot \sin(x) \). It outlines the selection of 'u' and 'dv', calculates 'du' and 'v', and applies the formula to find the integral. The explanation continues with the integral of \( x^2 \ln(x) \), emphasizing the importance of choosing 'u' and 'dv' wisely to simplify the integration process. The paragraph concludes with the final answer for the integral of \( x^2 \ln(x) \) and introduces additional examples to practice the technique.
📘 Advanced Integration by Parts Examples
This paragraph presents more complex examples of integration by parts, starting with the integral of \( x^2 e^x \). It discusses the strategy of choosing 'u' and 'dv' and the need for applying integration by parts multiple times due to the complexity of the function. The explanation proceeds with the integral of \( \ln(x) \) and \( x^2 \sin(x) \), demonstrating the process of repeatedly using integration by parts and simplifying the expressions to reach the final answers.
📙 Integration of Logarithmic and Exponential Functions
The fourth paragraph focuses on integrating functions involving logarithms and exponentials. It starts with the integral of \( \ln(x) \) using integration by parts and then moves on to the integral of \( x^2 e^x \), which requires a strategic approach to selecting 'u' and 'dv'. The explanation includes the step-by-step process of integrating these functions and concludes with the final expressions for each integral.
📒 Dealing with Higher Power Logarithms
This paragraph addresses the integration of higher power logarithmic functions, specifically \( \ln(x)^7 \). It explains how to rewrite the integral using logarithmic properties to simplify the process. The explanation demonstrates the integration of \( 7 \ln(x) \) and arrives at the final answer by applying the previously discussed anti-derivative of \( \ln(x) \).
📕 Complex Exponential and Trigonometric Integrals
The sixth paragraph tackles the integration of complex functions involving exponentials and trigonometric functions, such as \( e^x \sin(x) \). It describes the process of using integration by parts and the need for careful selection of 'u' and 'dv'. The explanation includes the iterative application of integration by parts and the algebraic manipulation required to simplify the integral to its final form.
📔 Integration of Products of Logarithms and Trigonometric Functions
The final paragraph discusses the integration of products of logarithmic and trigonometric functions, such as \( \ln(x)^2 / x \). It outlines the process of rewriting the integral and applying integration by parts, including the selection of 'u' and 'dv' and the calculation of 'du' and 'v'. The explanation concludes with the final answer for the integral and provides a method for solving similar problems.
📓 Definite Integrals with Exponential and Trigonometric Functions
This paragraph explores the calculation of a definite integral involving \( e^{3x} \cos(4x) \). It suggests making the trigonometric function the 'u' in the integration by parts process and outlines the steps to find the integral. The explanation includes the manipulation of constants and the application of integration by parts multiple times to arrive at the final expression for the integral.
Mindmap
Keywords
💡Indefinite Integral
💡Integration by Parts
💡Natural Logarithm (ln x)
💡Exponential Function (e^x)
💡Trigonometric Functions
💡Derivative
💡Antiderivatives
💡Chain Rule
💡Constant of Integration (C)
💡Power Rule
💡Product Rule
Highlights
Introduction to finding the indefinite integral of x times e^x using integration by parts.
Setting u=x and dv=e^x dx for the integral of x times e^x.
Deriving the integral of e^x as e^x and applying integration by parts formula.
Solving the integral of x times sine x dx using integration by parts.
Choosing u=x and dv=sin(x)dx for the integral of x times sine x.
Finding the antiderivative of sine x as -cosine x and applying it to the integral.
Integrating x squared ln x using a strategic choice of u and dv.
Deciding on u=ln(x) and dv=x^2 dx for the integral of x squared ln x.
Using the derivative of ln(x) as 1/x to simplify the integral.
Deriving the integral of ln(x) as x ln(x) - x and applying integration by parts.
Integrating x squared sine x dx using a two-step integration by parts process.
Choosing u=x^2 and dv=sin(x)dx for the integral of x squared sine x, and applying integration by parts twice.
Solving the integral of x squared e^x dx using a two-step integration by parts approach.
Deciding on u=x^2 and dv=e^x dx for the integral of x squared e^x, and applying integration by parts twice.
Finding the indefinite integral of ln(x) squared dx using integration by parts.
Rewriting ln(x) squared as 2ln(x) and applying integration by parts.
Solving the integral of ln(x) to the power of seven using properties of logarithms.
Integrating e^x sin(x) dx using a recursive integration by parts strategy.
Rewriting the integral of ln(x) squared over x as a product of two functions.
Solving the definite integral of e^(3x)cos(4x) dx using integration by parts.
Applying integration by parts to the integral of e^(3x)sin(4x) dx and simplifying the result.
Transcripts
so how can we find the indefinite
integral of x
times e raised to the x
to do this
we need to use something called
integration by parts
the integral of u
dv
is equal to u times v
minus the integral of v
d u
so we need to determine u
dv
v
and u and d u
let me put it in a different order
so which one should we set equal to u
and which one
should we set equal to dv
it's best to set u
equal to x
because if we do so
d u will simply be equal to one times dx
and when dealing with constants
they're a lot easier to integrate than
variables
so we're going to make dv equal to e to
the x dx
now v
is the integral of dv
the integral of e to the x is simply e
to the x
so now using the formula u times v
that's going to be x
times e raised to the x
minus the integral of v
d u
v
is e to the x d u is 1 dx
and the integral of
e to the x
is e to the x plus some constant c
so this is the final answer that's the
integral
of
x raised to the or x times e raised to
the x
let's try another problem
let's integrate x times sine x dx
so what should we make u
and
dv equal to
so just like before i'm going to make u
equal to x
so that d u is going to be 1
dx
now dv
has to be what remains
sine x
dx
now what is the integral of sine
the derivative of what function will
give us sign
the integral of sine is negative cosine
the derivative of negative cosine is
positive sine
so using the formula the integral of u
dv
is equal to u times v
minus the integral of v
d u
so u is x
v is negative cosine x
and then minus v
and d u is
1 dx
so far we have negative x
cosine x
and then these two negative signs will
become positive so plus
the integral of cosine x
dx
now the antiderivative of positive
cosine
is positive sine
so the final answer
is what you see here negative cosine i
mean negative x cosine x
plus sine x plus c
now here's another problem to work on
let's integrate x squared ln x
so what should we make u
and
dv equal to in this problem
if we make u equal to x squared
and we make dv equal to ln x
dx the integral of ln x
will be quite difficult to do
unless you know what it is so we're not
going to do that
so it's best to make
dv
x squared dx
and u is going to be ln x because we
know the derivative of l and x
the derivative of l and x
is 1 over x
and the antiderivative of x squared
the antiderivative of dv will give us v
so the antiderivative of x squared dx
is x cubed over 3.
so let's use the formula now
u times v that's going to be ln x times
x cubed over three
minus the integral of v
times d u
and
d u is
one over x
it's supposed to be dx
so this is equal to 1 3
x cubed times ln x
and then we could cancel an x and so we
have minus one third the integral of x
squared
dx
the integral of x squared is going to be
x to the third over three
and then plus some constant c
so the final answer is one third
x cubed ln x minus one over nine
x to the third plus c
and that concludes this problem
so let's work on some more examples
let's try this problem what is the
anti-derivative
of the natural log of x
how can we use integration by parts to
get this answer
now we can't make u
equal well we can't make a dv equal to l
and x
because we don't know the integral of l
and x we're trying to figure that out so
therefore we have to make u equal to
ln x the derivative of that is one over
x dx
dv
has to be
just we can make dv dx if we want to
or one dx
and so v the integral of one dx is just
x
so now let's start with the formula
and so it's going to be u times v
u is ln x
and v
is x
minus the integral of v
and then d u
is one over x
dx
so far we have x
ln x
x times one over x the x variables
cancel
so we have the integral of one
dx now the integral of one dx is x
so this is the final answer
x ln x
minus x plus c
that is the integral
of the natural log of x
let's find the indefinite integral of x
squared
sine x
dx
so which part should be u
and which part
is going to be
dv
we need to make u equal to x squared
and we need to differentiate it twice to
bring it down to a constant so this is
one of those problems where you have to
use integration
of you know integration by parts two
times
d u is going to be 2x dx
dv
is sine x dx
so this is dv
and this is going to be u
the integral of sine
is negative cosine
so let's start with the formula
so u times v that's x squared
times negative cosine x
minus
the integral of v which is negative
cosine x
times d u
that's
2x dx
so let's organize what we have negative
x squared cosine x
and then these two negative signs will
become positive so positive the integral
of
2x cosine x
we could take the 2 and move it to the
front so it's just going to be x cosine
x
now we need to use the integration by
parts formula
on the integral of x cosine x
we're going to make u equal to x
and
dv
is going to be cosine x
dx
the integral of cosine is sine and the
derivative of x is 1
dx
so using the formula
it's going to be u times v
which is
x times sine x
minus the integral
of v d u
so v is sine x
d u is just dx
and so the integral of sine
is
negative cosine
and then plus c
so now what we need to do is replace
this part
with what we have here
so it's going to be plus we need to
distribute 2 to everything in here
so it's going to be 2x
sine x
plus
this would be plus cosine x but times
two so plus two cosine x
and we don't need to multiply c by two
because c is just
some generic constant
and
this whole thing is equal to the
original integral and so this is the
answer
negative x squared cosine x
plus
2x sine x
plus 2 cosine x plus c
now let's work on this problem
let's say we have the integral of x
squared
e to the x dx
go ahead and try that problem
so let's begin by writing the formula
the integral of u
dv is equal to
u times v minus the integral of v
d u
now what should we make u equal to
and what should be
dv
i prefer making dv
e c x dx
because once we integrate it
it's going to stay the same
and i want x square to eventually become
a constant so this is one of those
problems where you need to do
use the integration by parts formula
twice
so we're going to make u
equal to x squared
du is going to be 2x dx
so using the formula it's going to be uv
that's going to be
x squared
that's you v is e to the x
minus the integral of v which is e to
the x
times d u that's 2x
dx
so now this becomes
well before we uh do that let's use
integration by parts one more time
so i'm going to make u
equal to 2x
du is going to be the derivative of 2x
which is 2 but times dx
dv
we're going to make that equal to
e x dx
just like we did before
and the integral of that is simply e x
so we're going to have
x squared
e x
and then minus
now let's use the formula again for this
part so it's uv
so 2x
e to the x
and then minus the integral
of vdu
so that's 2
e to the x
dx
or multiplying those two
now let's begin by distributing the
negative signs
so this is going to be negative 2x
e to the x and then these two negatives
will cancel
giving us
positive
well let's take out the two so positive
two integral
e to the x dx
so our final answer is going to be x
squared
e c x minus
2 x e to the x
anti derivative of e to the x is just e
to the x and then plus c
now if we want to we can factor out
e to the x
so we're going to have x squared minus
2x
plus 2 and then plus the constant c
so this right here is our final answer
that is the anti-derivative
of
x squared e to the x
what about this problem
what is the integral of
ln x
squared
dx
try that
so for this problem what we need to do
is we need to make u
equal
to ln x squared
dv is going to be equal to the other
part
we're going to make dv
equal to dx
so v
is going to be the integral of dx which
is x
d u i'm going to write that here i need
more space
we need to use the chain rule
so first let's use the power rule by
moving the 2 to the front
so it's going to be 2 and then we'll
keep everything on the inside the same 2
times ln x
and then we're going to subtract that by
1
and then take the derivative of the
inside the derivative of l and x is one
over x
so using the formula this is going to be
u times v
so v is x u is
ln x squared
minus the integral of v d u
v is x
d u is
2
ln x
times 1 of x and then
let's not forget the dx part
so we can cancel x and one of x
so this becomes x
ln x squared
and then we could take out the two we
can move it to the front so minus two
integral
ln x dx
earlier in this video we determined that
the integral of l and x is x ln x minus
x
so this is going to be -2
and then we can replace this with
x ln x minus x
now let's go ahead and distribute the
two so our final answer is going to be x
times ln x squared
minus 2x
ln x
and then negative 2 times negative x
that's going to be positive
2x and then plus c
so this
is
the indefinite integral
of ln x squared
now what would you do if you were to see
a problem like this
ln x to the seventh power dx
go ahead and try that
well first you need to realize that we
can rewrite this problem
a property of logs allows us to move the
exponent to the front
so make sure you understand this this is
not ln x
to the seventh power like this
if it was we can't move the seven
the seven is only affecting the x
variable as a result we can move the 7
to the front
so this becomes 7
ln x
so this is what we're integrating
this is equivalent to 7
integral
ln x dx
and we know the anti-derivative of l and
x it's x
ln x minus x
so this is our final answer it's just 7
x ln x
minus 7 x plus c
now let's try this one let's find the
indefinite integral
of e raised to the x sine x dx
go ahead and work on that
so for this problem i want to make
u equal to sine x
and then dv i'm going to set that equal
to
e to the x dx
so v is simply going to be e to the x
d u
is the derivative of sine x which is
cosine x
so using the formula it's going to be
uv
that's going to be e
sine x
minus the integral of v d u
so v is e to the x d u is cosine x dx
so we need to use integration by parts
one more time
but this time we're going to make
u equal to
cosine x instead of sine
and dv is going to be the same
dv is going to be e to the x
dx
if v is the same
d u is going to be the derivative of
cosine which is negative
sine
and then times dx
so now we're going to have
e x
sine x
minus and then in brackets
u v so that's going to be e x
cosine x
minus the integral of v to u
so v is e to the x d u is
negative
sine x and then dx
so let's go ahead and distribute the
negative sign i'm going to rewrite the
original integral
so the integral of e x sine x
that's going to be e to the x
sine x
these two negative signs
will become positive
and then we need to distribute this
negative so it's negative e raised to
the x cosine x
and then we need to apply the negative
to this
which is going to be negative
e to the x sine x
now the key to solving this problem is
to realize that this term is very
similar to
the original problem
so what we want to do is we want to
add this to both sides
if we move it to the other side we're
going to have 2
e x sine x
which is equal to this
now we want to find the value of just
one
integral e to the x sine x
so what we need to do is divide both
sides
by two
so that's we can say that our final
answer
is one half
we also factor out e to the x so it's
one half e to the x
and then times
sine x
minus cosine x
plus c
so that's the anti-derivative
of
e raised to the x times sine x
now let's move on to our next example
let's find the antiderivative of ln
x squared
divided by
x dx
so go ahead and try that
now the first thing that i want to do is
i'm going to rewrite this as
ln x squared times 1 over x
dx
i want to multiply as a product of two
functions
now what i'm going to do is i'm going to
make u
equal to ln x
squared
du will be 2
times ln x to the first power
times 1 over x
and then dx
so since i made u ln x squared
dv is going to be the other part
dv is going to be 1 over x
dx
so anti-derivative of 1 over x is ln x
now using the formula this is going to
be u times v
u is ln x
squared
v is just lnx
minus the integral of vdu
so v is ln x
d u is 2
ln x times 1 over x
dx
so let's put this all together so first
i'm going to rewrite the original
problem
that's equal to ln x squared times ln x
that's
ln x raised to the third power
in this 2 i'm going to move to the front
here we have ln x times ln x
that's ln x
squared
and it's multiplied by one over x which
we can move that to the bottom
and we need to realize that these two
terms are similar
so we need to add
this to both sides or if we move it to
the other side
we'll basically add in 1
plus 2 which will become 3. so on the
left side we're going to have 3
integral
ln x
squared
over x dx
and that's going to equal
ln x
to the third power
so if we divide both sides by 3
this 3 will disappear
and we're going to have it underneath
here
and then plus our constant c
so this
is the answer
to the original problem that's how we
can find the integral of ln x squared
over x
now for the sake of practice go ahead
and try this
finding a definite integral of e to the
3x cosine
4x dx
so whenever you have an e to the x times
the sine or cosine
it's good to make
the cosine part
equal to u
that's what i'd like to do personally
so in this one i'm going to make u equal
to cosine
4x
du
the derivative of cosine is going to be
negative
sine
4x and then times the derivative of the
angle the derivative of 4x is 4
and then we're going to have times dx
so dv in this problem
is going to be e
to the 3x to dx
the antiderivative of e to the 3x it's e
to the 3x over 3
which
we can write as one third e to the 3x
so using the formula
it's going to be uv
so that's 1 3
e to the 3x times cosine
4x
minus the integral
of v d u
so v
is
one third
e to the three x
d u
that's
negative
four
sine 4x
dx
so what we're going to do now is we're
going to move the constants to the front
the two negative signs
will cancel they will become positive
and we can move one third times four to
the front
so this is going to be plus
four over three
integral
e to the three x
sine four x dx
now we need to use integration by parts
one more time
so dv will be the same we don't need to
adjust this part of the formula
but u
is going to be sine 4x as opposed to
cosine for x
so du is going to change as well
the derivative of sine is cosine
4x
and then times 4 times dx
so using integration by parts we're
going to have u times v
so that's one-third
e to the 3x times sine
4x
and then minus
the integral of vdu
so v is one third
e to the three x
and then d u that's times four
cosine four x
d x
now let's go ahead and distribute four
over three
so multiply those together that's going
to be
4 over 9
e to the 3x
sine 4x
and then this 2 will also be 4 over 9
but it's going to be negative
four over nine
actually
there's a four here so this is four
thirds times four thirds
so that's 16 over nine with a negative
sign
and then we're going to have the
integral of e
to the 3x
cosine 4x
dx
now notice that these two are similar
so we're going to add
16 over 9 integral e to the 3x cosine
4xdx to both sides
so this is 1 but we can treat it as if
it's
9 over 9 just to get common denominators
so 9 over 9 plus
16 over 9
that's going to be
25
over 9
integral
e to the 3x
cosine 4x dx
and then that's going to be 1 3
e to the 3x cosine 4x
and then plus
4
over 9
e to the
3x sine 4x
now to get this to 1
what we're going to do is we're going to
multiply both sides by the reciprocal of
25 over nine
so that is by nine over twenty five
so those two fractions will cancel
on the left side we're just going to
have
e to the three x cosine four x
dx
and
what i'm going to do
i want to keep this in factored form
so we have 9 over twenty five
from one third and four over nine we can
factor out one third
and we can also factor out e to the
three x
so here we took out e to the three x
we're left with just cosine four x
here we took out e to the three x and we
took out one third from four over nine
so we're gonna have four thirds left
over times sine 4x
the last thing we could do is maybe
reduce this
so if we divide both numbers by 3 this
becomes
3
over 25
e to the 3x
and then times
cosine 4x plus 4 over 3
sine 4x plus c
so that's going to be the answer for
this problem
that's the indefinite integral of
e to the 3x cosine 4x
you
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