1. ## Integral of 1/(1+4x^2)

Problem:

$\displaystyle \int \frac{dx}{1+4x^2}dx$

Attempt:

$\displaystyle 4x = tan(\theta)$
$\displaystyle dx = \frac{d\theta}{cos^2(\theta)}$

$\displaystyle \int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

$\displaystyle \int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

$\displaystyle \int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

$\displaystyle \int d\theta$

$\displaystyle \theta + C$

$\displaystyle arctan(4x) + C$

$\displaystyle \frac{1}{2} arctan(2x) + C$

2. correction ...

$\displaystyle 1 + 4x^2 = 1 + (2x)^2$

$\displaystyle 2x = \tan{\theta}$

3. Originally Posted by Fallen186
Problem:

$\displaystyle \int \frac{dx}{1+4x^2}dx$

Attempt:

$\displaystyle 4x = tan(\theta)$
$\displaystyle dx = \frac{d\theta}{cos^2(\theta)}$

$\displaystyle \int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

$\displaystyle \int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

$\displaystyle \int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

$\displaystyle \int d\theta$

$\displaystyle \theta + C$

$\displaystyle arctan(4x) + C$
$\displaystyle \frac{1}{2} arctan(2x) + C$
Close. Try $\displaystyle 2x = \tan \theta$
4. If $\displaystyle 4x=\tan\,\theta$, then $\displaystyle x=\frac{1}{4}\tan\,\theta$ and $\displaystyle dx=\frac{1}{4\,\cos^2\theta}\,d\theta$.
5. $\displaystyle \int \frac{1}{1+4x^2}\, dx = \int \frac{1}{1+(2x)^{2}} \, dx = \frac{1}{2}\cdot arctan(2x)+C$