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Correct Answer is : xcot⁡−1x+log⁡1+x2+cx ^{-1} x + {1+x² + cxcot−1x+log1+x2+c

Correct Answer is : xcot⁡−1x+log⁡1+x2+cx ^{-1} x + {1+x² + cxcot−1x+log1+x2+c

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Indefinite Integration

Section: Mathematics

Question : 16 of 89

Marks: +1, -0

For

0 < x < 1, \int [\tan^{-1}(1-x+x^2) + \tan^{-1}(1-x)] \, dx =

Solution:

For

0 < x < 1

I = \int [\tan^{-1}(1-x+x^2) + \tan^{-1}(1-x)] \, dx

= \int \tan^{-1} \left[ \frac{(1-x+x^2)+(1-x)}{1-(1-x+x^2)(1-x)} \right] dx

= \int \tan^{-1} \left( \frac{2-2x+x^2}{x(2-2x+x^2)} \right) dx

= \int \tan^{-1} \left( \frac{1}{x} \right) dx

= \int \cot^{-1}(x) \, dx = \int \left( \frac{\pi}{2} - \tan^{-1} x \right) dx

= \int \frac{\pi}{2} \, dx - \int \tan^{-1} x \, dx

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

[Using product rule]

= \frac{\pi}{2} x - x \tan^{-1} x + \frac{1}{2} \ln (1+x^2) + C

where

C =

constant of integration

= x \left( \frac{\pi}{2} - \tan^{-1} x \right) + \frac{1}{2} \ln (1+x^2) + C

= x \cot^{-1} x + \frac{1}{2} \ln (1+x^2) + C

= x \cot^{-1} x + \log \sqrt{1+x^2} + C

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