Question

Integrate $\int \frac{1}{x^{2/3}\sqrt{x^{2/3}-4}}dx$

Answer 1

Solution:

$\int \frac{1}{x^{2/3}\sqrt{x^{2/3}-4}}dn$

We put $x^{2/3}-4=t^2$

$\frac{2}{3}.n^{-1/13}dn=2tdt$

$\frac{1}{3n^{1/3}}dn=tdt$

$\frac{dn}{x^{1/3}}=3tdt$

and, $x^{2/3}-4=t^2$

$x^{2/3}=t^2+4$

$x^{1/3}=\sqrt{t^2+4}$

$\int \frac{1}{x^{1/3}.x^{1/3}.\sqrt{x^{2/3}-4}}dn$

$\Rightarrow \int \frac{3t.dt}{\sqrt{t^2+4}.\sqrt{t^2}}$

$\Rightarrow \int \frac{3t.dt}{t.\sqrt{4+t^2}}$

$\Rightarrow 3\int \frac{dt}{\sqrt{4+t^2}}$

$\Rightarrow 3ln[t+\sqrt{4+t^2}]$

$\Rightarrow 3ln[\sqrt{x^{2/3}-4}+x^{1/3}]$

$=3ln[n^{1/3}+\sqrt{x^{2/3}-4}]$