Question

Integrate the function $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

Answer 1

$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}$
Let $x=t^6\Rightarrow dx=6t^{5}dt$
$\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx=\int \frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}dx$
$=\int \frac{6t^5}{t^2(1+t)}dt$
$=6\int \frac{t^3}{(1+t)}dt$
On dividing, we obtain
$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx=6\int \left\{\right(t^2-t+1)-\frac{1}{1+t}\}dt$
$=6[\left(\frac{t^3}{3}\right)-\left(\frac{t^2}{2}\right)+t-\log |1+t|]$
$=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log (1+x^{\frac{1}{6}})+C$
$=2\sqrt{x}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log (1+x^{\frac{1}{6}})+C$