Question

Integrate the following functions w.r.t. x.

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

Answer 1

$LetI=\int \frac{1}{x^{1/2}+x^{1/3}}dxPut{x}^{1/6}=t\Rightarrow x=t^6\Rightarrow dx=6t^{5}dt\therefore I=\int \frac{6t^5}{(t^6{)}^{1/2}+(t^6{)}^{1/3}}dt=6\int \frac{t^5}{t^3+t^2}dt=6\int \frac{t^5}{t^2(t+1)}dt=6\int \frac{t^3}{(t+1)}dt=6\int \left(\frac{t^3+1-1}{t+1}\right)dt=6\int [\frac{(t+1)(t^2-t+1)}{t+1}-\frac{1}{t+1}]dt=6\int \{t^2-t+1+\frac{-1}{t+1}\}dt=6\{\frac{t^3}{3}-\frac{t^2}{2}+t-log|t+1|\}+C=6\{\frac{(x^{1/6}{)}^3}{3}-\frac{(x^{1/6}{)}^2}{2}+(x^{1/6})-log|x^{1/6}+1|\}+C(\because t=x^{1/6})=2x^{1/2}-3x^{1/3}+6x^{1/6}-6log|x^{1/6}+1|+C$