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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.6
Integrate the functions in Exercises 134 to 155.
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Question : 141 of 271
Marks: +1, -0
xtan⁡−1xx \tan^{-1} x
Solution:  
Let I=∫xtan⁡−1x dxI = \int x \tan^{-1} x \, dx
=tan⁡−1x∫x dx−∫[(ddxtan⁡−1x)∫x dx]dx= \tan^{-1} x \int x \, dx - \int \left[ \left( \frac{d}{dx} \tan^{-1} x \right) \int x \, dx \right] dx
=tan⁡−1xx22−∫11+x2⋅x22 dx= \tan^{-1} x \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \, dx
=x22tan⁡−1x−12∫x2x2+1 dx= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx
=x22tan⁡−1x−12∫x2+1−1x2+1 dx= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2+1-1}{x^2+1} \, dx
=x22tan⁡−1x−12∫(1−11+x2)dx= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx
=x22tan⁡−1x−12(x−tan⁡−1x)+C= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \left( x - \tan^{-1} x \right) + C
=x22tan⁡−1x−12x+12tan⁡−1x+C= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} x + \frac{1}{2} \tan^{-1} x + C
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