# Show that

Question:

$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\left[\operatorname{Hint}: \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} \operatorname{Put} x=t^{6}\right]$

Solution:

$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$

Let $x=t^{6} \Rightarrow d x=6 t^{5} d t$

$\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=\int \frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} d x$

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

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

On dividing, we obtain

$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=6 \int\left\{\left(t^{2}-t+1\right)-\frac{1}{1+t}\right\} d t$

$=6\left[\left(\frac{t^{3}}{3}\right)-\left(\frac{t^{2}}{2}\right)+t-\log |1+t|\right]$

$=2 x^{\frac{1}{2}}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+\mathrm{C}$

$=2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+\mathrm{C}$