Question

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

Answer 1

$\Rightarrow LCM$ of $2$ and $3=6$

Thus $x=t^6$ equation $\left(1\right)$

$dx=6t^{5}dt$ equation $\left(2\right)$

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

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

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

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

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

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

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

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

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