### Topics

## Introduction to Indefinite Integrals - At A Glance:

## There's Always a Factor of 1

We can use integration by parts to find the integral of something that doesn't look like a product. This is because whatever the integrand is, we can think of it as the product of itself and 1. Then we can choose *v'* = 1 and apply the integration-by-parts formula.

For example, since

*ln* *x* = (*ln* *x*)(1),

we know

If we chose *u* = 1 then *u'* would be zero, which doesn't seem like a good idea. So take

*u* = *ln* *x*

*v'* = 1

## Factoring

Sometimes we need to rearrange the integrand in order to see what *u* and *v*' should be. Exponents can be deceiving.

### Sample Problem

For example, look at the integral

This looks like a product, so we want to use integration by parts. However, choosing

*u* = *x*^{5}

or

*u* = *e*^{x3}

won't work very well (try it yourself if you don't believe us; we're not going to demonstrate). But what if we rewrite the integrand by factoring *x*^{5}?

Now we can see it's reasonable to choose

*v'* = *x*^{2}*e*^{x3},

since we can use substitution to figure out the antiderivative *v*. This leaves

*u* = *x*^{3}.

## Integrating by Parts Twice

We already did some exercises where you had to integrate by parts twice: once to start off, then again to find the new integral. There are some problems where, if you integrate by parts twice, the original integral shows up again. Sometimes this will be as helpful as the equation

0 = 0

(that is, not helpful at all). However, sometimes when the original integral shows up again, you'll get an equation that you can rearrange to solve for the original integral.

#### Exercise 1

Apply the formula for integration by parts with *u* = *ln* x and *v'* = 1 to find

Answer

We take *u* and *v*' as suggested.

Then

Applying the formula,

Now there's magic. Rearranging the right-hand side, we get

which equals

*x ln* *x *– *x* + *C*.

So

#### Exercise 2

Use integration by parts to find .

Answer

It might be useful to remember that the derivative of arcsin x is

We can rewrite the integrand to show the factor of 1 that's already there:

We can't choose *v'* = arcsin *x*, because if we knew the antiderivative of arcsin *x* we'd already be done with this problem. So take

*u* = arcsin *x*

*v'* = 1

Then

We put everything in the formula and get

Thankfully, we can use substitution to find this new integral.

#### Exercise 3

Find

by applying the formula for integration by parts with *u* = *x*^{3} and *v'* = *x*^{2}*e*^{x3}.

Answer

If

*u* = *x*^{3}

*v'* =* x*^{2}*e*^{x3}

then we can use integration by substitution to find *v*.

Put all this stuff in the formula:

Happily, the integrand of the new integral is *v*' again, and we already know how to integrate that.

Here's the final answer:

#### Exercise 4

Find

Answer

Let's factor *x*^{7}.

Now the part

*x*^{3}cos(*x*^{4})

looks like the derivative of something, so

we can take

*u* = *x*^{4}

*v'* = *x*^{3}cos(*x*^{4}).

Then

Putting everything into the formula,

We can use substitution to find this new integral, so

#### Exercise 5

Use integration by parts to find

taking *u* = sin *x* for the first integration.

Answer

If

*u* = sin *x*

*v'* = *e*^{x}

then

*u' *= cos* x*

*v* = *e*^{x}

When we stuff everything into the integration-by-parts formula, we get

Now we need to use integration by parts again to find

Remembering the lessons of the previous example, we'll keep *u* and *v*' with their same parts by taking

*u* = cos *x*

*v'* = *e*^{x}

Then

*u'* = -sin *x*

*v* = *e*^{x}

We put this into the formula and get

Putting this back into the first application of the formula for gives us

(being careful to properly distribute the negative sign!).

Again, we've found a formula we can solve for .

#### Exercise 6

Integrate.

Answer

Take

*u* = *x*^{2}

*v'* = *e*^{2x}

so that *u'* will be simpler than *u*. Then

We apply the formula and get

Looks like we need to use integration by parts again to find

This is very much like our first example of integration by parts. Again, so that u' will be simpler than *u*, take

*u* = *x*

*v'* = *e*^{2x}

Then

We put everything in the formula and see what we get:

Wrap this up in parentheses (except for the + *C*), and put it back where we left off:

#### Exercise 7

Integrate.

Hint

You can find the derivative of arctan using the chain rule.

Answer

If you don't know the antiderivative of arctan *x* off the top of your head, that's to be expected. Thankfully there's a hidden factor of 1, so we can use integration by parts.Since we don't know the antiderivative of arctan *x* we can't use that for *v*', so we need to pick

*u* = arctan *x*

*v'* = 1

The derivative of arctan *x* is

which you could figure out using the chain rule, so

*u'* = (1 + *x*^{2})^{-1}

*v* = *x*.

Put everything in the formula:

Integrating the new integral by substitution (taking *u* = (1 + *x*^{2}) so *u*' = 2*x*) we get

#### Exercise 8

Integrate.

Answer

If you remember the antiderivative of *ln* *x* you could rewrite the integrand as

take both *u* and *v*' to be *ln* *x*, and go from there.

If you don't remember the antiderivative of *ln* *x* then you'll have to use the hidden factor of 1 and take

*u* = (*ln* *x*)^{2}

*v'* = 1.

Then

We put stuff in the formula:

Again, we use the hidden factor of 1 to get

Wrapping this expression in brackets and putting it in where we left off,

#### Exercise 9

Integrate.

Answer

We need to use the factoring trick on this one. If we break up *x*^{3} so that one *x* goes with (1 + *x*^{2})^{-3}, we'll have a nice candidate for *v*'.

Take

*u* = *x*^{2}

*v'* = *x*(1 + *x*^{2})^{-3}

Then

Putting this into the formula, we get

The new integral can be evaluated by substitution, taking *u* = (1 + *x*^{2}), and we get

#### Exercise 10

Integrate.

Answer

This one doesn't need any special tricks. Just take

*u* = (3*x* + 1)

*v' *= sin *x*

*u'* = 3

*v* = -cos *x*

and stick it all in the formula:

#### Exercise 11

Integrate.

Answer

This one doesn't need special tricks either. Take

and do the magic:

#### Exercise 12

Integrate.

Answer

Assuming you've forgotten the antiderivative of *ln* *x* again (we have), take

Do the magic: