The value of ₀¹ ^{-1}≤ft(1-x+x²) d x is

Correct Answer is : log⁡2 2log2

Correct Answer is : log⁡2 2log2

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MHTCET 23 Apr 2025 Shift 1 Paper

Question : 121 of 150

Marks: +1, -0

The value of

\int_{0}^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x

Solution:

Concept:

Use the identity

\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)

to split the integrand, then evaluate using integration by parts.

Explanation:

Let

I = \int_{0}^{1} \tan^{-1}\left(1-x+x^{2}\right)dx

Rewrite the argument as

1-x+x^{2} = \frac{1-x(1-x)}{x+(1-x)}

, since

x+(1-x)=1

Use the reciprocal identity

\tan^{-1}u + \tan^{-1}\left(\frac{1}{u}\right) = \frac{\pi}{2}

for

u>0

This gives

\tan^{-1}(1-x+x^{2}) = \frac{\pi}{2} - \tan^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right)

By the tangent addition formula,

\tan^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right) = \tan^{-1}x + \tan^{-1}(1-x)

I = \int_{0}^{1}\left[\frac{\pi}{2} - \tan^{-1}x - \tan^{-1}(1-x)\right]dx

Thus

I = \frac{\pi}{2} - \int_{0}^{1}\tan^{-1}x\,dx - \int_{0}^{1}\tan^{-1}(1-x)\,dx

Using

\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx

, the two integrals are equal.

Therefore,

I = \frac{\pi}{2} - 2\int_{0}^{1}\tan^{-1}x\,dx

Now integrate by parts:

\int_{0}^{1}\tan^{-1}x\,dx = \left[x\tan^{-1}x\right]_{0}^{1} - \int_{0}^{1}\frac{x}{1+x^{2}}dx

Evaluating,

\int_{0}^{1}\tan^{-1}x\,dx = \frac{\pi}{4} - \frac{1}{2}\left[\log(1+x^{2})\right]_{0}^{1} = \frac{\pi}{4} - \frac{1}{2}\log 2

Finally,

I = \frac{\pi}{2} - 2\left(\frac{\pi}{4} - \frac{1}{2}\log 2\right) = \log 2

Answer:

\log 2

, i.e., Option C.

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