Apply the formula for integration by parts with u = ln x and v' = 1 to find
We take u and v' as suggested.
Applying the formula,
Now there's magic. Rearranging the right-hand side, we get
x lnx – x + C.
Use integration by parts to find .
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
We put everything in the formula and get
Thankfully, we can use substitution to find this new integral.
by applying the formula for integration by parts with u = x3 and v' = x2ex3.
u = x3
v' = x2ex3
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:
Let's factor x7.
Now the part
looks like the derivative of something, so
we can take
u = x4
v' = x3cos(x4).
Putting everything into the formula,
We can use substitution to find this new integral, so
Use integration by parts to find
taking u = sin x for the first integration.
u = sin x
v' = ex
u' = cos x
v = ex
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' = ex
u' = -sin x
v = ex
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 .
u = x2
v' = e2x
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' = e2x
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:
You can find the derivative of arctan using the chain rule.
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
If you remember the antiderivative of lnx you could rewrite the integrand as
take both u and v' to be lnx, and go from there.
If you don't remember the antiderivative of lnx then you'll have to use the hidden factor of 1 and take
u = (lnx)2
v' = 1.
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,
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'.
u = x2
v' = x(1 + x2)-3
Putting this into the formula, we get
The new integral can be evaluated by substitution, taking u = (1 + x2), and we get
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:
This one doesn't need special tricks either. Take
and do the magic:
Assuming you've forgotten the antiderivative of lnx again (we have), take