Question
Integrate the function $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ [Hint: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$ Put x = t6]

Answer

Given: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ or we can write it as $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$ 
Let x = t$\Rightarrow$ dx = 6t5dt
$\Rightarrow \int \frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} \cdot d x=\int \frac{6 t^{5}}{t^{2}(1+t)} \cdot d t$ 
= $\text { 6. } \int \frac{t^{3}}{(1+t)} \cdot d t$ 
After division we get,
$\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ = $\int\left[\left(t^{2}-t+1\right)-\frac{1}{(1+t)}\right] \cdot d t$ 
= $6 .\left\{\int \mathrm{t}^{2} . \mathrm{dt}-\int \mathrm{t} . \mathrm{dt}+\int 1 . \mathrm{dt}-\int\left[\frac{1}{(1+t)}\right] \cdot \mathrm{dt}\right\}$ 
= $6\left[\left(\frac{t^{3}}{3}\right)-\left(\frac{t^{2}}{2}\right)+t-\log (1+t)\right]$ 
= $6\left[\left(\frac{\left(x^{\frac{1}{6}}\right)^{3}}{3}\right)-\left(\frac{\left(x^{\frac{1}{6}}\right)^{2}}{2}\right)+\left(x^{\frac{1}{6}}\right)-\log \left(1+\left(x^{\frac{1}{6}}\right)\right)\right]+c$ 
= $\left[\left(2 x^{\frac{1}{2}}\right)-\left(3 x^{\frac{1}{3}}\right)+6 \cdot x^{\frac{1}{6}}-6 \cdot \log \left(1+x^{\frac{1}{6}}\right)\right]+C$ 
= $2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+c$ 

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