Question

evaluate integral dx/1+3x^2​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{dx}{1 + 3 {x}^{2} }$

$Let \: x = \frac{1}{ \sqrt{3} } \tan( \alpha ) \\ \implies dx = \frac{1}{ \sqrt{3} } \sec^{2} ( \alpha ) d \alpha$

$= \int\frac{ \frac{1}{ \sqrt{3} } \sec^{2} ( \alpha ) }{1 + 3. \frac{1}{3} \tan^{2} ( \alpha ) } d \alpha$

$= \frac{1}{ \sqrt{3} } \int\frac{ \sec^{2} ( \alpha ) }{1 + \tan^{2} ( \alpha ) } d \alpha$

$= \frac{1}{ \sqrt{3} } \int\frac{ \sec^{2} ( \alpha ) }{ \sec^{2} ( \alpha ) } d \alpha$

$= \frac{1}{ \sqrt{3} } \int d \alpha$

$= \frac{1}{ \sqrt{3} } \ \alpha + c$

$= \frac{1}{ \sqrt{3} } \tan ^{ - 1} ( \sqrt{3}x ) + c$