# Simple antiderivatives

• Nov 30th 2008, 10:56 PM
DemonGal711
Simple antiderivatives
Okay, I'm doing Integral tests for series and I think my brain is just off cause I can't find these simple antiderivatives. My first thought was make it an negative exponent, but yeah, that doesn't really help anything. Can someone just explain one to me cause I can't even think how to start these?

anti 1/(2n - 1)
anti 1/(n^2 + 4)
• Dec 1st 2008, 12:45 AM
janvdl
Quote:

Originally Posted by DemonGal711
Okay, I'm doing Integral tests for series and I think my brain is just off cause I can't find these simple antiderivatives. My first thought was make it an negative exponent, but yeah, that doesn't really help anything. Can someone just explain one to me cause I can't even think how to start these?

anti 1/(2n - 1)
anti 1/(n^2 + 4)

The first one:

We will use u-substitution.

$\displaystyle \int \frac{1}{2n-1} \ dn$

--
$\displaystyle u = 2n - 1$
$\displaystyle du = 2n \rightarrow \frac{1}{2} \ du = dn$
--

$\displaystyle \frac{1}{2} \int \frac{1}{u} \ du$

$\displaystyle = \frac{1}{2} \ln{u} + C$

$\displaystyle = \frac{1}{2} \ln{2n - 1} + C$

_____________

The second one:

$\displaystyle \int \frac{1}{4 + n^2} \ dn$

You'll have to study the forms of these integrals so you can easily spot this one is going to be arctan.

For an integral of this form:
$\displaystyle \int \frac{1}{a^2 + u^2} \ du = \frac{1}{a} tan^{-1} \left( \frac{u}{a} \right) + C$

So in your case we have:

$\displaystyle \int \frac{1}{4 + n^2} \ dn = \frac{1}{2} tan^{-1} \left( \frac{n}{2} \right) + C$