Choose the correct answer in Exercise: The value of ^ 1 _ 0 ^ -1 … - Vidyadip

MCQ

Choose the correct answer in Exercise:
The value of $\int^{1}_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$ is

A
1
✓
0
C
-1
D
$\frac{\pi}{4}$

✓

Answer

Correct option: B.

0

$\text{Let I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{\text{x}-(1-\text{x})}{1+\text{x}(1-\text{x})}\bigg)\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}\text{x}-\tan^{-1}(1-\text{x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{\text{1}}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}(1-1+\text{x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{1}_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$

Adding (1) and (2), we obtain

$\Rightarrow2\text{I}=\int^{1}\limits_{0}\Big(\tan^{-1}\text{x)}+\tan^{-1}(1-\text{x)}-\tan^{-1}\text{x}\Big)\text{dx}$

$\Rightarrow2\text{I}=0$

$\Rightarrow\text{I}=0$

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