# Thread: what is antideriv. of arctan?

1. ## what is antideriv. of arctan?

hey,

I've got this problem:

[indefinite integral](1/(t^3+t))dt

I've used the partial fractions to split the integral into 2 separate parts, which becomes:

ln|t| + [indefinite integral](t/(t^2 + 1))dt + C

here I used the integration by parts for the bolded integral, which gives me:

ln|t| + t*arctan(t) + [definite integral](arctan(t))dt + C

and I cannot figure out what is the antiderivative of arctan(t) is?

thanks for the tips,
Tuugii

2. $\displaystyle \int \frac{1}{t}-\frac{t}{t^2+1}=\int \frac{1}{t}-\frac{1}{2}\int \frac{(t^2+1)'}{t^2+1}=ln|t|-\frac{1}{2}ln|t^2+1|$

no arctan involved just regualr subtitution

3. i don't know how to solve arctg, but why didn't u use a subtitution
$\displaystyle t^2+1=u$
That would make the integral solvable.

4. As the previous posts say, dealing with arctangent is not an issue here. however, if you are still curious, we would find that integral using integration by parts, with $\displaystyle u = \arctan x$ and $\displaystyle dv = 1$

5. Originally Posted by Tuugii
I've used the partial fractions to split the integral into 2 separate parts,
you dont even need to use that for this question, check this out

$\displaystyle \int \frac{1}{t(t^2+1)}= \frac{1+t^2-t^2}{t(t^2+1)}= \int \frac{1+t^2}{t(t^2+1)}-\frac{t^2}{t(t^2+1)}=\int \frac{1}{t}-\frac{t}{t^2+1}$

6. thanks a lot for the replies.