Q266 / 270 · NDA 2015 (Phase 1) · Mathematics
What is $\int \frac{d x}{\sqrt{x^2+a^2}}$ equal to?
where $c$ is the constant of integration.
- Correct answerA$\ln \left|\frac{\mathrm{x}+\sqrt{\mathrm{x}^2+a^2}}{a}\right|+c$
- B$\ln \left|\frac{x-\sqrt{x^2+a^2}}{a}\right|+c$
- C$\ln \left|\frac{x^2+\sqrt{x^2+a^2}}{a}\right|+c$
- DNone of these
Solution
$\begin{aligned} & \int \frac{d x}{\sqrt{x^2+a^2}} \\ & \text {Let } x=a \tan u \\ & \quad d x=a \sec ^2 u d u \\ & =\int \frac{a \sec ^2 u d u}{\sqrt{a^2 \tan ^2 u+a^2}} \\ & =a \int \frac{\sec ^2 u d u}{\sqrt{a^2\left(1+\tan ^2 u\right)}} \Rightarrow \frac{a}{a} \int \frac{\sec ^2 u d u}{\sec u}\end{aligned}$
$\begin{aligned} & =\int \frac{\sec ^2 u d u}{\sec u}=\int \sec u d u \\ & =\ln [\tan (u)+\sec (u)]+c \quad\left[\because \int \sec x d x=\tan x\right]\end{aligned}$
$\begin{aligned}
& =\ln \left[\frac{x}{\sqrt{a^2}}+\sqrt{1+\frac{x^2}{a^2}}\right]+c \\
& =\ln \left[\frac{x}{a}+\frac{\sqrt{a^2+x^2}}{a}\right]+c \\
& =\ln \left[\frac{x+\sqrt{x^2+a^2}}{a}\right]+c
\end{aligned}$
$\therefore \quad$ Option (A) is correct.
Hence, the correct answer is option: 'A'.