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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2019 · Mathematics · Indefinite Integration

What is ∫ (d x)/(x(1+ x)^n) equal to (n^1 1) ?

What is ∫dxx(1+ln⁡x)n\int \frac{d x}{x(1+\ln x)^n} equal to (n11)\left(n^1 1\right) ?

Options

  1. A1(n−1)(1+ln⁡x)n+1+c\frac{1}{(n-1)(1+\ln x)^{n+1}}+c
  2. B1−n(1+ln⁡x)n−1+c\frac{1-n}{(1+\ln x)^{n-1}}+c
  3. Cn+1(1+ln⁡x)n−1+c\frac{n+1}{(1+\ln x)^{n-1}}+c
  4. D1(n−1)(1+ℓnx)n−1+c\frac{1}{(n-1)(1+\ell n x)^{n-1}}+c

Correct answer

D. 1(n−1)(1+ℓnx)n−1+c\frac{1}{(n-1)(1+\ell n x)^{n-1}}+c

Step-by-step solution

∫dxx(1+ℓnx)nLet 1+ℓnx=t⇒1xdx=dt⇒∫dttn=t−n+1−n+1\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}=(1+ℓnx)−n+11−n=1(n−1)(1+ℓnx)n−1+C\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}$
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