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

Problem:

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

Attempt:

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

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

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

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

$\int d\theta$

$\theta + C$

$arctan(4x) + C$

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

2. correction ...

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

$2x = \tan{\theta}$

3. Originally Posted by Fallen186
Problem:

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

Attempt:

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

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

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

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

$\int d\theta$

$\theta + C$

$arctan(4x) + C$

$
\frac{1}{2} arctan(2x) + C
$
Close. Try $2x = \tan \theta$

4. If $4x=\tan\,\theta$, then $x=\frac{1}{4}\tan\,\theta$ and $dx=\frac{1}{4\,\cos^2\theta}\,d\theta$.

5. $\int \frac{1}{1+4x^2}\, dx = \int \frac{1}{1+(2x)^{2}} \, dx = \frac{1}{2}\cdot arctan(2x)+C$