Q112 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\begin{aligned} & \int \frac{x d x}{\sqrt[15]{\left(1+x^2\right)^{12}\left(2+x^2\right)^{18}}}=\alpha\left(\frac{1+x^2}{2+x^2}\right)^{1 / n}+C \Rightarrow \\ & \frac{n}{\alpha}=\end{aligned}$
- A$6$
- B$4$
- Correct answerC$2$
- D$8$
Solution
$\begin{aligned} I & =\int \frac{x d x}{\sqrt[15]{\left(1+x^2\right)^{12}\left(2+x^2\right)^{18}}} \\ & =\int \frac{x d x}{\left(1+x^2\right)^{12 / 15}\left(2+x^2\right)^{18 / 15}} \\ & =\int \frac{x d x}{\left(1+x^2\right)^{4 / 5}\left(2+x^2\right)^{6 / 5}} \\ & =\int \frac{x d x}{\left(1+x^2\right)^2\left(\frac{2+x^2}{1+x^2}\right)^{6 / 5}}\end{aligned}$
Put $u=\frac{2+x^2}{1+x^2}$
$\begin{aligned} & \Rightarrow \quad d u=\left\{\frac{\left(1+x^2\right)(2 x)-\left(2+x^2\right)(2 x)}{\left(1+x^2\right)^2}\right\} d x \\ & \Rightarrow d u=\frac{2 x\left(1+x^2-2-x^2\right) d x}{\left(1+x^2\right)^2} \\ & \Rightarrow d u=\frac{-2 x d x}{\left(1+x^2\right)^2} \\ & \Rightarrow \quad \frac{x d x}{\left(1+x^2\right)^2}=-\frac{d u}{2}\end{aligned}$
Then, $I=-\frac{1}{2} \int \frac{d x}{(u)^{6 / 5}}$
$\begin{aligned} & =-\frac{1}{2}\left\{\frac{u^{\frac{-6}{5}}+1}{\left(-\frac{6}{5}+1\right)}\right\}+C \\ & =\frac{5}{2} \times(u)^{-1 / 5}+C \\ & =\frac{5}{2}\left(\frac{1+x^2}{2+x^2}\right)^{1 / 5}+C\end{aligned}$
Thus, $\alpha=\frac{5}{2}$ and $n=5$
Then, $\frac{n}{\alpha}=\frac{5}{5 / 2}=2$
Hence, the correct answer is option: 'C'.