Q110 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Mathematics

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

  • Correct answer
    A
    $\frac{\pi}{2} x+C$
  • B
    $\frac{\pi}{4} x+C$
  • C
    $x+C$
  • D
    $\pi x+C$

Solution

$I=\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1} x d x+\int \tan ^{-1}(1-x) d x$ $\begin{aligned} & =\int\left[\tan ^{-1}\left(1-x+x^2\right) d x+\tan ^{-1}\left(\frac{x+1-x}{1-x(1-x)}\right)\right] d x \\ & =\int\left[\tan ^{-1}\left(1-x+x^2\right)+\tan ^{-1}\left(\frac{1}{1-x+x^2}\right)\right] \mathrm{dx} \\ & =\int\left[\tan ^{-1}\left(1-x+x^2\right) d x+\cot ^{-1}\left(1-x+x^2\right)\right] d x \\ & =\int \frac{\pi}{2} d x=\frac{\pi}{2} x+C\end{aligned}$

Hence, the correct answer is option: 'A'.

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