1. Integration problem

I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is:

Simplifying further, I get:

and

But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it.

2. $\displaystyle \int \dfrac{1}{b^2+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C$

3. Originally Posted by TwoPlusTwo
I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is:

Simplifying further, I get:

and

But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it.

$\displaystyle{\int\frac{1}{2+x^2}\,dx=\frac{1}{\sq rt{2}}\int\frac{1/\sqrt{2}\,dx}{1+\left(\frac{x}{\sqrt{2}}\right)^2} }$ $\display{=\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{ \sqrt{2}}\right)+C}$

Tonio

4. Originally Posted by harish21
$\displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C$
That holds if and only if $b = 1$.

$\displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k$.

5. Originally Posted by TheCoffeeMachine
That holds if and only if $b = 1$.

$\displaystyle \int \dfrainc{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k$.
indeed. it's usually written $\displaystyle \int \frac 1{b^2 + x^2}~dx = \frac 1b \tan ^{-1} \frac xb + C$

using $\displaystyle a$ instead of $\displaystyle b$