Test Index

UPSC NDA II 2014 Mathematics Paper

Show Para Hide Para

For the next two (02) items that follow:

Consider

\int x \tan^{-1} x \, dx = A (x^2+1) \tan^{-1} x + B x + C

where C is the constant of integration.

[NDA II 2014]

© examsnet.com

Question : 80 of 120

Marks: +1, -0

What is the value of B?

1
1/2
-1/2
1/4

Solution:

\int\limits x\tan^{-1}x\,dx = A(x^2+1)\tan^{-1}x+Bx+C

\Rightarrow \tan^{-1}x \int\limits x\,dx - \int\limits \frac{1}{1+x^2} \frac{x^2}{2}\,dx

\Rightarrow \frac{x^2}{2} \tan^{-1}x - \frac{1}{2} \int\limits \frac{x^2}{1+x^2}\,dx

\Rightarrow \frac{x^2}{2} \tan^{-1}x - \frac{1}{2} \int\limits \left(1 - \frac{1}{1+x^2}\right)\,dx + C

\Rightarrow \frac{x^2}{2} \tan^{-1}x - \frac{x}{2} + \frac{1}{2} \tan^{-1}x + C

\Rightarrow \frac{1}{2} \left[ \tan^{-1}x (x^2+1) - x \right] + C

\Rightarrow A = \frac{1}{2}

and

B = -\frac{1}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question