The value of $\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x$ is

The value of $\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x$ is
  1. 2
  2. -1
  3. 1
  4. 0

Solution

$\begin{aligned} & \text { Let } I=\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x \\ & =\int_0^1 \tan ^{-1}\left[\frac{x+(x-1)}{1+x(1-x)}\right] d x=\int_0^1 \tan ^{-1}\left[\frac{x+(x-1)}{1-(x-1)(x)}\right] d x \\ & =\int_0^1\left[\tan ^{-1} x+\tan ^{-1}(x-1) d x=\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1}(x-1) d x\right. \\ & \left.=\left[\left[x \tan ^{-1} x\right]_0^1-\frac{1}{2} \int_0^1 \frac{2 x}{1+x^2} d x\right]+\left[x \tan ^{-1}(x-1)\right]_0^1-\frac{1}{2} \int_0^1 \frac{2 x-2+2}{1+(x-1)^2} d x\right] \\ & =\left(\frac{\pi}{4}\right)-\frac{1}{2}\left[\log \left|1+x^2\right|\right]_0^1+0-\frac{1}{2} \int_0^1 \frac{2 x-2}{1+(x-1)^2} d x-\int_0^1 \frac{d x}{1+(x-1)^2} \\ & =\left(\frac{\pi}{4}\right)-\frac{1}{2} \log 2-\frac{1}{2}\left[\log \left|1+(x-1)^2\right|\right]_0^1-\left[\tan ^{-1}(x-1)\right]_0^1 \\ & =\frac{\pi}{4}-\frac{1}{2} \log 2-\frac{1}{2}(0-\log 2)-\left[0-\tan ^{-1}(-1)\right] \\ & =\frac{\pi}{4}-\frac{1}{2} \log 2+\frac{1}{2} \log 2-\frac{\pi}{4}=0\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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