Q232 / 270 · NDA 2019 (Phase 2) · Mathematics

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

  • A
    $\frac{1}{(n-1)(1+\ln x)^{n+1}}+c$
  • B
    $\frac{1-n}{(1+\ln x)^{n-1}}+c$
  • C
    $\frac{n+1}{(1+\ln x)^{n-1}}+c$
  • Correct answer
    D
    $\frac{1}{(n-1)(1+\ell n x)^{n-1}}+c$

Solution

$\begin{aligned} & \int \frac{d x}{x(1+\ell \mathrm{n} x)^n} \\ & \text {Let } 1+\ell \mathrm{n} x=t \\ & \Rightarrow \frac{1}{x} d x=d t \\ & \Rightarrow \int \frac{d t}{t^n}=\frac{t^{-n}+1}{-n+1}\end{aligned}$ $\begin{aligned} & =\frac{(1+\ell n x)^{-n+1}}{1-n} \\ & =\frac{1}{(n-1)(1+\ell \mathrm{n} x)^{n-1}}+C\end{aligned}$

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

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