The value of ^1_0 ^ -1 ( 2x-1 1+x-x^2 ) dx, is: - Vidyadip

MCQ

The value of $\int\limits^1_0\tan^{-1}\Big(\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\Big)\text{ dx},$ is:

A
$1$
✓
$0$
C
$-1$
D
$\frac{\pi}{4}$

✓

Answer

Correct option: B.

$0$

Let, $\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{i})$
$=\int\limits^1_0\tan^{-1}\frac{2(1-\text{x})-1}{1+(1-\text{x})-(1-\text{x})^2}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{2-\text{x}-1-\text{x}^2+2\text{x}}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{1+\text{x}-\text{x}^2}\text{ dx}$
$=-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{ii})$
Adding $(i)$ and $(ii)$
$2\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}$
Hence, $\text{I}=0$

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