Q104 / 180 · COMEDK 2017 · Mathematics

The value of $\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}}$ is

  • A
    $3 \sqrt{x}+3(\sqrt[3]{x})-6 \sqrt[6]{x}+\log (\sqrt[6]{x}+1)+C$
  • B
    $2 \sqrt{x}+6(\sqrt[6]{x})-6 \log (\sqrt[6]{x}+1)+C$
  • Correct answer
    C
    $2 \sqrt{x}-3(\sqrt[3]{x})+6 \sqrt[6]{x}-6 \log (\sqrt[6]{x}+1)+C$
  • D
    None of the above

Solution

$\operatorname{Let} I=\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}}$ Put $x=t^{6} \Rightarrow d x=6 t^{5} d t$ $\begin{aligned} \therefore \quad I &=\int \frac{6 t^{5}}{t^{3}+t^{2}} d t=6 \int \frac{t^{3}}{t+1} d t \\ &=6 \int\left(t^{2}-t+1-\frac{1}{t+1}\right) d t \\ &=2 t^{3}-3 t^{2}+6 t-6 \log (t+1)+C \\ &=2 \sqrt{x}-3(\sqrt[3]{x})+6(\sqrt[6]{x})-6 \log (\sqrt[6]{x}+1)+C \end{aligned}$

Hence, the correct answer is option: 'C'.

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