Q213 / 270 · NDA 2021 (Phase 2) · Mathematics

What is $\int \frac{d x}{x\left(x^2+1\right)}$ equal to?

  • Correct answer
    A
    $\frac{1}{2} \ln \left(\frac{x^2}{x^2+1}\right)+C$
  • B
    $\ln \left(\frac{x^2}{x^2+1}\right)+C$
  • C
    $\frac{3}{2} \ln \left(\frac{x^2}{x^2+1}\right)+C$
  • D
    $\frac{1}{2} \ln \left(\frac{x^2+1}{x^2}\right)+C$

Solution

$\begin{aligned} & \mathrm{I}=\int \frac{d x}{x\left(x^2+1\right)} \\ & =\int \frac{d x}{x\left(x^2+1\right)} \\ & \text {Let } x^2+1=t \Rightarrow x^2=t-1 \\ & 2 x d s=d t \\ & =\frac{1}{2} \int \frac{d t}{t(t-1)} \\ & =\frac{-1}{2} \int\left[\frac{1}{t}-\frac{1}{(t-1)}\right] d t \\ & =\frac{-1}{2}\left[\int \frac{1}{t} d t-\int \frac{1}{(t-1)} d t\right] \\ & =\frac{-1}{2}[\log (t)-\log (t-1)]+C \\ & =\frac{1}{2}[\log (t-1)-\log t]+C\end{aligned}$ $\begin{aligned} & =\frac{1}{2} \log \left(\frac{t-1}{t}\right)+C \\ & =\frac{1}{2} \log \left(\frac{x^2}{x^2+1}\right)+C\end{aligned}$

Hence, the correct answer is option: 'A'.

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