Indefinite Integrals Exercises

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

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.

Answer

If

u = x³

v' = x²e^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:

Answer

Let's factor x⁷.

Now the part

x³cos(x⁴)

looks like the derivative of something, so

we can take

u = x⁴

v' = x³cos(x⁴).

Then

Putting everything into the formula,

We can use substitution to find this new integral, so

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 .

Answer

Take

u = x²

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:

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 + x²)^-1

v = x.

Put everything in the formula:

Integrating the new integral by substitution (taking u = (1 + x²) so u' = 2x) we get

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 we'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,

Answer

We need to use the factoring trick on this one. If we break up x³ so that one x goes with (1 + x²)^-3, we'll have a nice candidate for v'.

Take

u = x²

v' = x(1 + x²)^-3

Then

Putting this into the formula, we get

The new integral can be evaluated by substitution, taking u = (1 + x²), and we get

Answer

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

u = (3x + 1)

v' = sin x

u' = 3

v = -cos x

and stick it all in the formula:

Answer

This one doesn't need special tricks either. Take

and do the magic:

Answer

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

Let integration by parts do its magic: