# 7. Integration by Parts

by M. Bourne

Sometimes we meet an integration that is the product of 2 functions. We may be able to integrate such products by using Integration by Parts.

If u and v are functions of x, the product rule for differentiation that we met earlier gives us:

d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)

Rearranging, we have:

u(dv)/(dx)=d/(dx)(uv)-v(du)/(dx)

Integrating throughout with respect to x, we obtain the formula for integration by parts:

This formula allows us to turn a complicated integral into more simple ones. We must make sure we choose u and dv carefully.

NOTE: The function u is chosen so that (du)/(dx) is simpler than u.

Continues below

## Priorities for Choosingu

When you have a mix of functions in the expression to be integrated, use the following for your choice of u, in order.

1. Let u = ln x

2. Let u = x^n

3. Let u = e^(nx)

### Example 1

intx\ sin 2x\ dx

#### Solution

We need to choose u. In this question we don't have any of the functions suggested in the "priorities" list above.

We could let u = x or u = sin 2x, but usually only one of them will work. In general, we choose the one that allows (du)/(dx) to be of a simpler form than u.

So for this example, we choose u = x and so dv will be the "rest" of the integral, dv = sin 2x dx.

We have u = x so du = dx.

Also dv = sin 2x\ dx and integrating gives:

v=intsin 2x\ dx

=(-cos 2x)/2

Substituting these 4 expressions into the integration by parts formula, we get (using color-coding so it's easier to see where things come from):

int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \   =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}

int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sin 2x dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:{-cos2x}/2:}} - int \color{blue}{\fbox{:{-cos2x}/2:}\ \color{magenta}{\fbox{:dx:}}

 = (-xcos2x)/2 + 1/2 int cos2x dx

 = (-xcos2x)/2 + 1/2 (sin2x)/2 +K

 = (-xcos2x)/2 + (sin2x)/4 +K

If the above is a little hard to follow (because of the line breaks), here it is again in a different format:

### Example 2

int x sqrt(x+1) dx

intxsqrt(x+1)\ dx

We could let u=x or u=sqrt(x+1).

Once again, we choose the one that allows (du)/(dx) to be of a simpler form than u, so we choose u=x.

Therefore du = dx. With this choice, dv must be the "rest" of the integral: dv=sqrt(x+1)\ dx.

u = x so du=dx.

dv=sqrt(x+1)\ dx, and integrating gives:

v=intsqrt(x+1) dx

=int(x+1)^(1//2)dx

=2/3(x+1)^(3//2)

Substituting into the integration by parts formula, we get:

int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \   =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}

int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sqrt(x+1) dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:2/3(x+1)^(3//2):}}  - int \color{blue}{\fbox{:2/3(x+1)^(3//2):}\ \color{magenta}{\fbox{:dx:}}

 = (2x)/3(x+1)^(3//2) - 2/3 int (x+1)^{3//2}dx

 = (2x)/3(x+1)^(3//2)  - 2/3(2/5) (x+1)^{5//2} +K

 = (2x)/3(x+1)^(3//2)- 4/15(x+1)^{5//2} +K

Once again, here it is again in a different format:

### Example 3

intx^2 ln 4x\ dx

intx^2ln\ 4x\ dx

We could let u=x^2 or u=ln\ 4x..

Considering the priorities given above, we choose u = ln\ 4x and so dv will be the rest of the expression to be integrated dv = x^2\ dx.

With u=ln\ 4x, we have du=dx/x.

Integrating dv = x^2\ dx gives:

v=intx^2dx=x^3/3

Substituting in the Integration by Parts formula, we get:

int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \   =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}

int \color{green}{\fbox{:x^2:}}\ \color{red}{\fbox{:ln 4x dx:}} = \color{green}{\fbox{:ln 4x:}}\ \color{blue}{\fbox{:x^3/3:}}  - int \color{blue}{\fbox{:x^3/3:}\ \color{magenta}{\fbox{:dx/x:}}

 = (x^3 ln 4x)/3 - 1/3 int x^2 dx

 = (x^3 ln 4x)/3  - 1/3 x^3/3 +K

 = (x^3 ln 4x)/3 - x^3/9 +K

Once again, here it is again in a different format:

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### Example 4

intx\ sec^2 x\ dx

int x\ sec^2 x\ dx

We choose u=x (since it will give us a simpler du) and this gives us du=dx.

Then dv will be dv=sec^2x\ dx and integrating this gives v=tan x.

Substituting these into the Integration by Parts formula gives:

int x\ sec^2 x\ dx =intu\ dv

=uv-intv\ du

=(x)(tan x)-int(tan x)dx

=x\ tan x-(-ln\ |cos x|)+K

=x\ tan x+ln\ |cos x|+K

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### Example 5

intx^2e^(-x)dx

intx^2 e^-x dx

The 2nd and 3rd "priorities" for choosing u given earlier said:

2. Let u = x^n

3. Let u = e^(nx)

This questions has both a power of x and an exponential expression. But we choose u=x^2 as it has a higher priority than the exponential. (You could try it the other way round, with u=e^-x to see for yourself why it doesn't work.)

So u=x^2 and this gives du=2x\ dx.

That leaves dv=e^-x\ dx and integrating this gives us v=-e^-x.

We substitute these into the Integration by Parts formula to give:

intx^2 e^-x dx =intu\ dv

=uv-intv\ du

=x^2(-e^-x)-int(-e^-x)(2x\ dx)

=-x^2e^-x+2intxe^-x dx

Now, the integral we are left with cannot be found immediately. We need to perform integration by parts again, for this new integral.

This time we choose u=x giving du=dx.

Once again we will have dv=e^-x\ dx and integrating this gives us v=-e^-x.

Substituting into the integration by parts formula gives:

int x e^-x dx =intu\ dv

=uv-intv\ du

=x(-e^-x)-int(-e^-x)dx

=-xe^-x+inte^-x dx

=-xe^-x-e^-x

So putting this answer together with the answer for the first part, we have the final solution:

intx^2e^-xdx =-x^2e^-x+2(-xe^-x-e^-x)

=-e^-x(x^2+2x+2)+K

### Example 6

int ln x dx

int ln\ x\ dx

Our priorities list above tells us to choose the logarithm expression for u. (of course, there's no other choice here. :-)

So with u=ln\ x, we have du=dx/x.

Then dv will simply be dv=dx and integrating this gives v=x.

Subsituting these into the Integration by Parts formula gives:

int ln\ x\ dx=int u\ dv

=uv-intv\ du

=x\ ln\ x-intx(dx)/x

=x\ ln\ x-intdx

=x\ ln\ x-x+K

### Example 7

intarcsin x dx

Using integration by parts, we set:

u=arcsin x, giving du=1/sqrt(1-x^2)dx.

Then dv=dx and integrating gives us v=x.

We now use:

intu\ dv=uv-intv\ du

This gives us:

int arcsin x\ dx =x\ arcsin x-intx/(sqrt(1-x^2))dx

Now, for that remaining integral, we just use a substitution (I'll use p for the substitution since we are using u in this question already):

p = 1 − x^2

So dp=-2x\ dx

This will yield:

intx/(sqrt(1-x^2))dx =-1/2int(dp)/sqrtp

=-1/2(2sqrtp)+K

=-sqrt(1-x^2)+K

int arcsin x\ dx =x\ arcsin x-(-sqrt(1-x^2))+K

= \x\ arcsin x+sqrt(1-x^2)+K

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This time we integrated an inverse trigonometric function (as opposed to the earlier type where we obtained inverse trigonometric functions in our answer). See Integration: Inverse Trigonometric Forms.

### Alternate Method for Integration by Parts

Here's an alternative method for problems that can be done using Integration by Parts. You may find it easier to follow.

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