Q112 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\int \frac{1}{\left(1+x^2\right) \sqrt{x^2+2}} d x=$
- Correct answerA$-\tan ^{-1} \frac{\sqrt{x^2+2}}{|x|}+c$
- B$-\tan ^{-1} \sqrt{x^2+2}+c$
- C$\tan ^{-1} \sqrt{\frac{x^2+1}{x^2+2}}+c$
- D$-\tan ^{-1} \sqrt{\frac{x^2+2}{x^2+1}}+c$
Solution
$I=\int \frac{1}{\left(1+x^2\right) \sqrt{x^2+2}} d x$
Let $x=\sqrt{2} \tan u \Rightarrow d x=\sqrt{2} \sec ^2 u d v$ $\begin{aligned} & \therefore \mathrm{I}=\sqrt{2} \int \frac{\sec ^2 u}{\left(2 \tan ^2 u+1\right) \sqrt{2 \tan ^2 u+2}} d u \\ & \mathrm{I}=\int \frac{\sec u}{2 \tan ^2 u+1} d u \end{aligned}$
$\mathrm{I}=\int-\cot u \operatorname{cosec} u\left(-\frac{1}{\operatorname{cosec}^2 u+1}\right) d u$
Let $\mathrm{v}=\operatorname{cosec} \mathrm{u} \Rightarrow d v=-\cot u \operatorname{cosec} u d u$ $\begin{aligned} & \therefore \mathrm{I}=-\int \frac{1}{v^2+1} d v=-\tan ^{-1} v+c \\ & \mathrm{I}=-\tan ^{-1}(\operatorname{cosec} u)+\mathrm{c}=-\tan ^{-1} \frac{\sqrt{x^2+2}}{|x|}+c \end{aligned}$
Let $x=\sqrt{2} \tan u \Rightarrow d x=\sqrt{2} \sec ^2 u d v$ $\begin{aligned} & \therefore \mathrm{I}=\sqrt{2} \int \frac{\sec ^2 u}{\left(2 \tan ^2 u+1\right) \sqrt{2 \tan ^2 u+2}} d u \\ & \mathrm{I}=\int \frac{\sec u}{2 \tan ^2 u+1} d u \end{aligned}$
$\mathrm{I}=\int-\cot u \operatorname{cosec} u\left(-\frac{1}{\operatorname{cosec}^2 u+1}\right) d u$
Let $\mathrm{v}=\operatorname{cosec} \mathrm{u} \Rightarrow d v=-\cot u \operatorname{cosec} u d u$ $\begin{aligned} & \therefore \mathrm{I}=-\int \frac{1}{v^2+1} d v=-\tan ^{-1} v+c \\ & \mathrm{I}=-\tan ^{-1}(\operatorname{cosec} u)+\mathrm{c}=-\tan ^{-1} \frac{\sqrt{x^2+2}}{|x|}+c \end{aligned}$
Hence, the correct answer is option: 'A'.