Question

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

Answer 1

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

Taking L.C.M of denominator of$\frac{1}{2}$ and $\frac{1}{3}$

which is 6.

So $x=t^6$

$\Rightarrow dx=6t^5$

Now

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

$=\int \frac{6t^5}{t^2(t+1)}dt$

$=\int \frac{6t^3}{t+1}dt$

$=\int \frac{6\left[\right(t^3+1)-1]}{t+1}dt$

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

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

$=6[\frac{\sqrt{x}}{3}-\frac{x^{1/3}}{2}+x^{1/6}-log/x^{1/6}+1]+C$