$\int \frac{\log \left(x^2+\mathrm{a}^2\right)}{x^2} \mathrm{~d} x=$

$\int \frac{\log \left(x^2+\mathrm{a}^2\right)}{x^2} \mathrm{~d} x=$
  1. $\frac{-\log \left(x^2+\mathrm{a}^2\right)}{x}+\frac{1}{\mathrm{a}} \tan ^{-1} \frac{x}{\mathrm{a}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  2. $\frac{-\log \left(x^2+\mathrm{a}^2\right)}{x}+\frac{2}{\mathrm{a}} \tan ^{-1} \frac{x}{\mathrm{a}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  3. $\frac{\log \left(x^2+\mathrm{a}^2\right)}{x^2}-\frac{1}{\mathrm{a}} \tan ^{-1} \frac{x}{\mathrm{a}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\frac{\log \left(x^2+\mathrm{a}^2\right)}{x^2}-\frac{2}{\mathrm{a}} \tan ^{-1} \frac{x}{\mathrm{a}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

Let $\begin{aligned} \mathrm{I} & =\int \frac{\log \left(x^2+\mathrm{a}^2\right)}{x^2} \mathrm{~d} x \\ & =\int \log \left(x^2+\mathrm{a}^2\right) \cdot x^{-2} \mathrm{~d} x \\ & =\log \left(x^2+\mathrm{a}^2\right) \int x^{-2} \mathrm{~d} x \\ & \quad-\int\left\{\frac{\mathrm{d}}{\mathrm{d} x}\left[\log \left(x^2+\mathrm{a}^2\right)\right] \int x^{-2} \mathrm{~d} x\right\} \mathrm{d} x \\ & =\log \left(x^2+\mathrm{a}^2\right) \cdot\left(-\frac{1}{x}\right)-\int \frac{2 x}{x^2+\mathrm{a}^2} \cdot\left(-\frac{1}{x}\right) \mathrm{d} x \\ & =-\frac{\log \left(x^2+\mathrm{a}^2\right)}{x}+2 \int \frac{1}{x^2+\mathrm{a}^2} \mathrm{~d} x \\ & =-\frac{\log \left(x^2+\mathrm{a}^2\right)}{x}+\frac{2}{\mathrm{a}} \tan ^{-1}\left(\frac{x}{\mathrm{a}}\right)+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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