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The value of $\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx$ is
A. 1
B. 0
C. -1
D. $\dfrac{\pi }{4}$

Last updated date: 14th Jul 2024
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Hint: Approach with one of the properties of definite integrals,
$\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}}$. Then use the identity \[{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x\], to get the required result.

Complete step-by-step answer:
We are given the following expressions,
$\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx$…… (1)

Let the equation (1) be considered to be I.
Then it can be written as,
$I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx........(2)$

To solve and get the value of ‘I’ we have to use the following identity,
$\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}}$
So by using the identity we have to replace x by (1+0-x)=(1-x) in the expression of I.

Thus we get,
$I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2-2x-1}{1+(1-x)-{{(1-x)}^{2}}} \right)}dx$
So, now doing the calculations carefully we get,
$I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{1-2x}{2-x-1+2x-{{x}^{2}}} \right)}dx$
$I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{1-2x}{1+x-{{x}^{2}}} \right)}dx$

Now taking out the minus sign, we get
$I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( -\dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx$
Now to simplify it more we will use the identity,
\[{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x\]

So now using the above identity, the integral ‘I’ can be written as,
$I=-\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx$
Substituting the value from equation (2), we get
Now bringing ‘I’ on one side, which becomes,
So, I=0

Therefore, the value of $\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)}dx=0.$
Hence, the correct answer is option B.

Note: In this type of question, students should be careful while using the identities and in the calculation part. Another approach is to first calculate the integration using integrate by parts method, and then applying the limits. You will get the same answer but that is a very lengthy and tedious process.