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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2014 · Mathematics · Indefinite Integration

Consider ∫ x ^ -1 x d x=A(x^2+1) ^ -1 x+B x+C , where C is the...

Consider ∫xtan⁡−1xdx=A(x2+1)tan⁡−1x+Bx+C\int x \tan ^{-1} x d x=\mathrm{A}\left(x^2+1\right) \tan ^{-1} x+\mathrm{B} x+\mathrm{C}, where C is the constant of integration. What is the value of B?

Options

  1. A1
  2. B12\frac{1}{2}
  3. C−12-\frac{1}{2}
  4. D14\frac{1}{4}

Correct answer

C. −12-\frac{1}{2}

Step-by-step solution

∫x⋅tan⁡−1xdx=A(x2+1)tan⁡−1x+Bx+CLet I=∫x⋅tan⁡−1xdx=tan⁡−1x⋅(x22)−∫11+x2⋅x22dx=x22⋅tan⁡−1x−12∫x21+x2dx=x22⋅tan⁡−1x−12∫(1+x2)−11+x2dx=x22⋅tan⁡−1x−12[∫1dx−∫11+x2dx]=x22⋅tan⁡−1x−12[x−tan⁡−1x]+C=x22⋅tan⁡−1x−12x+12tan⁡−1x+C=(x2+12)tan⁡−1x−x2+C\begin{aligned} & \int x \cdot \tan ^{-1} x d x=\mathrm{A}\left(x^2+1\right) \tan ^{-1} x+\mathrm{B} x+\mathrm{C} \\ & \text{Let } \mathrm{I}=\int x \cdot \tan ^{-1} x d x \\ &=\tan ^{-1} x \cdot\left(\frac{x^2}{2}\right)-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x \\ &=\frac{x^2}{2} \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} d x \\ &=\frac{x^2}{2} \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{\left(1+x^2\right)-1}{1+x^2} d x \\ &=\frac{x^2}{2} \cdot \tan ^{-1} x-\frac{1}{2}\left[\int 1 d x-\int \frac{1}{1+x^2} d x\right] \\ &=\frac{x^2}{2} \cdot \tan ^{-1} x-\frac{1}{2}\left[x-\tan ^{-1} x\right]+\mathrm{C} \\ &=\frac{x^2}{2} \cdot \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+\mathrm{C} \\ &=\left(\frac{x^2+1}{2}\right) \tan ^{-1} x-\frac{x}{2}+\mathrm{C}\end{aligned} Value of B=−12\mathrm{B}=-\frac{1}{2}
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