Evaluate $\displaystyle \int \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} \; dx$
Let $I = \displaystyle \int \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} \; dx$ $\;\;\; \cdots$ (1)
Now, $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$
Multiply and divide equation (1) with $x^{5/6}$
$\therefore$ $\;$ We have $I = \displaystyle \int \dfrac{1}{x^{5/6}} \times \dfrac{x^{1/2} \times x^{5/6} }{\left(\sqrt{x} + \sqrt[3]{x}\right)} \; dx$
i.e. $I = \displaystyle \int \dfrac{x^{-5/6} \times x^{4/3} \; dx}{x^{1/2} + x^{1/3}}$ $\;\;\; \cdots$ (1a)
Let $x^{1/6} = u$ $\;\;\; \cdots$ (2a)
Differentiating equation (2a) gives
$\dfrac{1}{6} \; x^{-5/6} \; dx = du$ $\implies$ $x^{-5/6} \; dx = 6 du$ $\;\;\; \cdots$ (2b)
From equation (2a),
$x^{1/2} = \left(x^{1/6}\right)^{3} = u^3$ $\;\;\; \cdots$ (2c)
$x^{1/3} = \left(x^{1/6}\right)^{2} = u^2$ $\;\;\; \cdots$ (2d)
$x^{4/3} = \left(x^{1/3}\right)^{4} = \left(u^2\right)^4 = u^8$ $\;\;\; \cdots$ (2e)
$\therefore$ $\;$ In view of equations (2a), (2b), (2c), (2d) and (2e), equation (1) can be written as
$\begin{aligned}
I & = \int \dfrac{6 \; u^8 \; du}{u^3 + u^2} \\\\
& = \int \dfrac{6 \; u^6 \; du}{u + 1} \;\;\; \cdots (3)
\end{aligned}$
$\begin{array}{rll}
u + 1 ) & u^6 & (u^5 - u^4 + u^3 - u^2 + u - 1 \\
& \underline{u^6 + u^5} & \\
& \hspace{6mm} -u^5 & \\
& \hspace{6mm} \underline{-u^5 - u^4} & \\
& \hspace{15mm} u^4 & \\
& \hspace{15mm} \underline{u^4 + u^3} & \\
& \hspace{21mm} -u^3 & \\
& \hspace{21mm} \underline{-u^3 - u^2} & \\
& \hspace{30mm} u^2 & \\
& \hspace{30mm} \underline{u^2 + u} & \\
& \hspace{36mm} -u & \\
& \hspace{36mm} \underline{- u - 1} & \\
& \hspace{43.5mm} 1 &
\end{array}$
$\therefore$ $\;$ $\dfrac{u^6}{u + 1} = u^5 - u^4 + u^3 - u^2 + u - 1 + \dfrac{1}{u + 1}$ $\;\;\; \cdots$ (4)
$\therefore$ $\;$ We have from equations (3) and (4),
$\begin{aligned}
I & = 6 \left\{\int u^5 \; du - \int u^4 \; du + \int u^3 \; du - \int u^2 \; du + \int u \; du - \int du + \int \dfrac{du}{u + 1} \right\} \\\\
& = 6 \left\{\dfrac{u^6}{6} - \dfrac{u^5}{5} + \dfrac{u^4}{4} - \dfrac{u^3}{3} + \dfrac{u^2}{2} - u + \log \left|u + 1\right| + c \right\} \\\\
& = x - \dfrac{6}{5} \; x^{5/6} + \dfrac{3}{2} \; x^{2/3} - 2 \; x^{1/2} + 3 \; x^{1/3} + 6 \log \left|x^{1/6} + 1\right| + c \\\\
& \left[\text{In view of equation (2a)}\right]
\end{aligned}$