# How do I find the integral intarctan(4x)dx ?

Mar 6, 2018

$I = x \cdot {\tan}^{-} 1 \left(4 x\right) - \frac{1}{4} \log | \sqrt{1 + 16 {x}^{2}} | + C$
$= x \cdot {\tan}^{-} 1 \left(4 x\right) - \frac{1}{8} \log | \left(1 + 16 {x}^{2}\right) | + C$

#### Explanation:

$\left(1\right) I = \int {\tan}^{-} 1 \left(4 x\right) \mathrm{dx}$
Let, ${\tan}^{-} 1 \left(4 x\right) = u \Rightarrow 4 x = \tan u \Rightarrow 4 \mathrm{dx} = {\sec}^{2} u \mathrm{du}$$\Rightarrow \mathrm{dx} = \frac{1}{4} {\sec}^{2} u \mathrm{du}$
$I = \int u \cdot \frac{1}{4} {\sec}^{2} u \mathrm{du} = \frac{1}{4} \int u \cdot {\sec}^{2} u \mathrm{du}$
Using Integration by Parts, $I = \frac{1}{4} \left[u \cdot \int {\sec}^{2} u \mathrm{du} - \int \left(\frac{d}{\mathrm{du}} \left(u\right) \cdot \int {\sec}^{2} u \mathrm{du}\right) \mathrm{du}\right] = \frac{1}{4} \left[u \cdot \tan u - \int 1 \cdot \tan u \mathrm{du}\right]$$= \frac{1}{4} \left[u \cdot \tan u - \log | \sec u |\right] + C$=1/4[tan^-1(4x)*(4x)-log|sqrt(1+tan^2u|]+C$= x \cdot {\tan}^{-} 1 \left(4 x\right) - \frac{1}{4} \log | \sqrt{1 + 16 {x}^{2}} | + C$
Second Method:
$\left(2\right) I = \int 1 \cdot {\tan}^{-} 1 \left(4 x\right) \mathrm{dx}$$= {\tan}^{-} 1 \left(4 x\right) \cdot x - \int \left(\frac{1}{1 + 16 {x}^{2}} \cdot 4\right) x \mathrm{dx}$
$= x \cdot {\tan}^{-} 1 \left(4 x\right) - \frac{1}{8} \int \frac{32 x}{1 + 16 {x}^{2}} \mathrm{dx}$
$= x \cdot {\tan}^{-} 1 \left(4 x\right) - \frac{1}{8} \log | 1 + 16 {x}^{2} | + C$