Question

Evaluate: $\int \frac{1}{ \sqrt{3x^{2}+8}}\ dx$

Answer 1

$ln( {x}^{n} ) dx = {x}^{n + 1} \div n + 1$
$ln( {(3 {x}^{2} + 8) }^{ \frac{ - 1}{2} } )dx = \frac{( {3 {x}^{2} + 8)}^{ \frac{ - 1}{2} + 1} }{ \frac{ - 1}{2} + 1}$
$= \frac{ {(3 {x}^{2} + 8)}^{ \frac{1}{2} } }{ \frac{1}{2} }$

$2 \sqrt{3 {x}^{2} + 8} + c$
I think so it's correct