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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2023 · Paper 2 · Mathematics · Indefinite Integration

∫ ^ -1 x dx is equal to

∫tan⁡−1xdx\int \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx} is equal to

Options

  1. A(x+1)tan⁡−1x−x+C(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C
  2. Bxtan⁡−1x−x+Cx \tan ^{-1} \sqrt{x}-\sqrt{x}+C
  3. Cx−xtan⁡−1x+C\sqrt{x}-x \tan ^{-1} \sqrt{x}+C
  4. Dx−(x+1)tan⁡−1x+C\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C

Correct answer

A. (x+1)tan⁡−1x−x+C(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C

Step-by-step solution

We have, I=∫1⋅tan⁡−1xdxI=\int 1 \cdot \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}
Using by parts,
I=tan⁡−1x⋅(x)−∫11+x×12x×xdx=xtan⁡−1x⋅−∫x(1+x)2xdx=xtan⁡−1x−∫(1+x(1+x)2x−1(1+x)2x)dx=xtan⁡−1x−∫dx2x+∫dx2x(1+x)=xtan⁡−1x−x+tan⁡−1x+C=(x+1)tan⁡−1x−x+C\begin{aligned} \\ I & =\tan ^{-1} \sqrt{x} \cdot(x)-\int \frac{1}{1+x} \times \frac{1}{2 \sqrt{x}} \times x d x \\ \\ & =x \tan ^{-1} \sqrt{x} \cdot-\int \frac{x}{(1+x) 2 \sqrt{x}} d x \\ \\ & =x \tan ^{-1} \sqrt{x}-\int\left(\frac{1+x}{(1+x) 2 \sqrt{x}}-\frac{1}{(1+x) 2 \sqrt{x}}\right) d x \\ \\ & =x \tan ^{-1} \sqrt{x}-\int \frac{d x}{2 \sqrt{x}}+\int \frac{d x}{2 \sqrt{x}(1+x)} \\ \\ & =x \tan ^{-1} \sqrt{x}-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\ \\ & =(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C \\ \end{aligned}
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