Question

Evaluate:
$\int \frac{1}{x^{1/2}+x^{1/3}}dx$.

Answer 1

Let $x=t^6$, $\Rightarrow dx=6t^{5}dt$

$\Rightarrow I=\int \frac{6t^{5}dt}{t^3+t^2}=6\int \frac{t^3}{1+t}dt$

$=6\int \frac{t(t^2-1+1)dt}{t+1}=6\int t(t-1)+\frac{t}{t+1}dt$

$=6\int t^2-t+1-\frac{1}{t+1}dt$

$=6[\frac{t^3}{3}-\frac{t^2}{2}+t-ln|t+1|]$

$=6[\frac{x^{1/2}}{3}-\frac{x^{1/3}}{2}+x^{1/6}-ln|1+x^{1/6}|]+c$.