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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Definite Integration

NDA 2014 · Mathematics · Definite Integration

What is ∫_0^1 e^ ^ -1 x d x 1+x^2 equal to?

What is ∫01etan⁡−1xdx1+x2\int_0^1 \frac{e^{\tan ^{-1} x} d x}{1+x^2} equal to?

Options

  1. Aeπ4−1e^{\frac{\pi}{4}}-1
  2. Beπ4+1e^{\frac{\pi}{4}}+1
  3. Ce−1e-1
  4. Dee

Correct answer

A. eπ4−1e^{\frac{\pi}{4}}-1

Step-by-step solution

∫01etan⁡−1x1+x2dx\int_0^1 \frac{e^{\tan ^{-1} x}}{1+x^2} d x Substituting t=tan⁡−1x⇒dtdx=11+x2⇒dt=dx1+x2\quad t=\tan ^{-1} x\Rightarrow \quad \frac{d t}{d x}=\frac{1}{1+x^2}\Rightarrow \quad d t=\frac{d x}{1+x^2} when x=0,t=tan⁡−1=0\begin{aligned} x & =0, t=\tan ^{-1} \\ & =0 \end{aligned} for x=1,t=tan⁡−1(1)=π4⇒∫01etan⁡−1x1+x2dx=∫0π4etdt=(et)0π4=eπ4−1\quad x=1, t=\tan ^{-1}(1)=\frac{\pi}{4}\Rightarrow \quad \int_0^1 \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int_0^{\frac{\pi}{4}} e^t d t=\left(e^t\right)_0^{\frac{\pi}{4}}=e^{\frac{\pi}{4}}-1
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