Question

Evaluate:

$\int \frac{x}{\sqrt{3x^2+4}}dx$.

Answer 1

Now,

$\int \frac{x}{\sqrt{3x^2+4}}dx$

$=\frac{1}{6}\int \frac{6x}{\sqrt{3x^2+4}}dx$

$=\frac{1}{6}\int \frac{d(3x^2+4)}{\sqrt{3x^2+4}}dx$

$=\frac{1}{6}.2.\sqrt{3x^2+4}+c$ [ Where $c$ is integrating constant]

$=\frac{1}{3}\sqrt{3x^2+4}+c$