# Math Help - Simple antiderivatives

1. ## 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)

2. 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.

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

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

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

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

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

_____________

The second one:

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

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:
$\int \frac{1}{a^2 + u^2} \ du = \frac{1}{a} tan^{-1} \left( \frac{u}{a} \right) + C$

So in your case we have:

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