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NCERT Class XII Mathematics Chapter - - Solutions

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Choose the correct answers in Exercises 269 to 272.
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Question : 272 of 271
Marks: +1, -0
The value of ∫01tan⁡−1(2x−11+x−x2)dx\int\limits_{0}^{1}\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx is
Solution:  
Let I=∫01tan⁡−1(2x−11+x−x2)dxI=\int\limits_{0}^{1}\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx
=∫01tan⁡−1[x+x−11−x(x−1)]dx=\int\limits_{0}^{1}\tan^{-1}\left[\frac{x+x-1}{1-x(x-1)}\right]dx
=∫01[tan⁡−1x+tan⁡−1(x−1)]dx=\int\limits_{0}^{1}\left[\tan^{-1}x +\tan^{-1}(x-1)\right]dx
=∫01tan⁡−1x dx+∫01tan⁡−1(x−1) dx=\int\limits_{0}^{1}\tan^{-1}x\,dx +\int\limits_{0}^{1}\tan^{-1}(x-1)\,dx
=I1+I2=I_1+I_2....(i)
Integrating by parts
I=tan⁡−1x⋅x∣01−(∫01x1+x2dx+tan⁡−1(x−1)⋅x∣01)−∫01x1+(x−1)2dxI = \tan^{-1}x \cdot x \Big|_0^1 - \left( \int\limits_{0}^{1}\frac{x}{1+x^2}dx + \tan^{-1}(x-1)\cdot x \Big|_0^1 \right) - \int\limits_{0}^{1}\frac{x}{1+(x-1)^2}dx
=(π4−12log⁡(1+x2))∣01+0−∫01xx2+1−2x+1dx=\left(\frac{\pi}{4}-\frac{1}{2}\log(1+x^2)\right)\Big|_0^1+0-\int\limits_{0}^{1}\frac{x}{x^2+1-2x+1}dx
=π4−12log⁡2+0−∫01xx2−2x+2dx=\frac{\pi}{4}-\frac{1}{2}\log 2+0-\int\limits_{0}^{1}\frac{x}{x^2-2x+2}dx
=π4−12log⁡2−12∫012x−2+2x2−2x+2dx=\frac{\pi}{4}-\frac{1}{2}\log 2-\frac{1}{2}\int\limits_{0}^{1}\frac{2x-2+2}{x^2-2x+2}dx
=π4−12log⁡2−12∣log⁡(x2−2x+2)∣01−12∫012x2−2x+2dx=\frac{\pi}{4}-\frac{1}{2}\log 2-\frac{1}{2}\left|\log(x^2-2x+2)\right|_0^1-\frac{1}{2}\int\limits_{0}^{1}\frac{2}{x^2-2x+2}dx
=π4−12log⁡2−12log⁡1+12log⁡2−∫011x2−2x+2dx=\frac{\pi}{4}-\frac{1}{2}\log 2-\frac{1}{2}\log 1+\frac{1}{2}\log 2-\int\limits_{0}^{1}\frac{1}{x^2-2x+2}dx
=π4−∫011(x−1)2+12dx=\frac{\pi}{4}-\int\limits_{0}^{1}\frac{1}{(x-1)^2+1^2}dx
=(π4−tan⁡−1(x−1))∣01=\left(\frac{\pi}{4}-\tan^{-1}(x-1)\right)\Big|_0^1
=π4−0+tan⁡−1(−1)=\frac{\pi}{4}-0+\tan^{-1}(-1)
=π4−π4=0=\frac{\pi}{4}-\frac{\pi}{4}=0
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