$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=$

$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=$
  1. $\sqrt{x}-\sqrt[3]{x}+\sqrt[6]{x}-\log |\sqrt[6]{x}+1|+c$
  2. $2 \sqrt{x}-3 \sqrt[3]{x}+6 \sqrt[6]{x}-6 \log |\sqrt[6]{x}+1|+c$
  3. $2 \sqrt{x}+3 \sqrt[3]{x}+6 \sqrt[6]{x}+6 \log |\sqrt[6]{x}+1|+c$
  4. $\sqrt{x}+\sqrt[3]{x}+\sqrt[6]{x}+\log |\sqrt[6]{x}+1|+c$

Solution

Let $I=\int \frac{d x}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ Put $x^{\frac{1}{6}}=t \Rightarrow x=t^6 \Rightarrow d x=6 t^5$ dt. Also $x^{\frac{1}{2}}=t^3$ and $x^{\frac{1}{3}}=t^2$ $\begin{aligned} & I=6 \int \frac{t^5 d t}{t^2(1+t)} \\ & =6 \int \frac{t^3}{(1+t)} d t=6 \int \frac{\left(t^3+1\right)-1}{(1+t)}=6 \int \frac{(t+1)\left(t^2-t+1\right)}{(1+t)} d t-6 \int \frac{d t}{1+t} \\ & =6 \int\left(t^2-t+1\right) d t-6 \log |1+t|+c=\frac{6 t^3}{3}-\frac{6 t^2}{2}+6 t-6[\log |1+t|]+c \\ & =2 \sqrt{x}-3 \sqrt[3]{x}+6 \sqrt[6]{x}-6 \log |1+\sqrt[6]{x}|+c \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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