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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2013 · Mathematics · Indefinite Integration

What is ∫ d x 4+x^2 equal to? where c is an arbitrary constant.

What is ∫dx4+x2\int \frac{d x}{\sqrt{4+x^2}} equal to? where c is an arbitrary constant.

Options

  1. Aln⁡∣4+x2+x∣+c\ln \left|\sqrt{4+x^2}+x\right|+c
  2. Bln⁡∣4+x2−x∣+c\ln \left|\sqrt{4+x^2}-x\right|+c
  3. Csin⁡−1(x2)+c\sin ^{-1}\left(\frac{x}{2}\right)+c
  4. DNone of the above

Correct answer

A. ln⁡∣4+x2+x∣+c\ln \left|\sqrt{4+x^2}+x\right|+c

Step-by-step solution

I=∫dxx2+4=∫dxx2+(2)2 Let x=2tan⁡θdx=2sec⁡2θdθ∫2sec⁡2θ⋅dθ22(1+tan⁡2θ)=∫sec⁡θ⋅dθ=ln⁡(sec⁡θ+tan⁡θ)+C=ln⁡(x+x2+4)+C\begin{aligned} & \mathrm{I}=\int \frac{d x}{\sqrt{x^2+4}}=\int \frac{d x}{\sqrt{x^2+(2)^2}} \\ & \text{ Let } \quad x=2 \tan \theta \\ & d x=2 \sec ^2 \theta d \theta \\ & \int \frac{2 \sec ^2 \theta \cdot d \theta}{\sqrt{2^2\left(1+\tan ^2 \theta\right)}}=\int \sec \theta \cdot d \theta \\ & =\ln (\sec \theta+\tan \theta)+C=\ln \left(x+\sqrt{x^2+4}\right)+C \end{aligned}
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