If 2\int\limits_0^1 \tan ^ - 1xdx = \int\limits_0^1 \cot ^ - 1 ( 1 - x + x^2 )dx, then \int\limits_0^1 \tan ^ - 1 ( 1 - x + x^2 )dx is equalto :

Definite Integration PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q42 with Solution

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Correct answer: C

Let $I=\int_0^1 \tan^{-1}(1-x+x^2)\,dx.$ We are given $2\int_0^1 \tan^{-1}x\,dx=\int_0^1 \cot^{-1}(1-x+x^2)\,dx.$
Use the standard identity for positive arguments: $\tan^{-1}t+\cot^{-1}t=\frac{\pi}{2}\quad (t>0).$ Here, $1-x+x^2=x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\quad \text{for }x\in[0,1].$ So, $\cot^{-1}(1-x+x^2)=\frac{\pi}{2}-\tan^{-1}(1-x+x^2).$ Hence,

=\frac{\pi}{2}-I.$$ 3. Therefore the given condition becomes $$2\int_0^1 \tan^{-1}x\,dx=\frac{\pi}{2}-I,$$ so $$I=\frac{\pi}{2}-2\int_0^1 \tan^{-1}x\,dx.$$ 4. Now evaluate $$\int_0^1 \tan^{-1}x\,dx.$$ Using integration by parts: Let $$u=\tan^{-1}x,\qquad dv=dx.$$ Then $$du=\frac{dx}{1+x^2},\qquad v=x.$$ Thus $$\int \tan^{-1}x\,dx=x\tan^{-1}x-\int \frac{x}{1+x^2}\,dx =x\tan^{-1}x-\frac12\ln(1+x^2).$$ So, $$\int_0^1 \tan^{-1}x\,dx=\left[x\tan^{-1}x-\frac12\ln(1+x^2)\right]_0^1 =\frac{\pi}{4}-\frac12\ln 2.$$ 5. Substitute into the expression for $I$: $$I=\frac{\pi}{2}-2\left(\frac{\pi}{4}-\frac12\ln2\right) =\frac{\pi}{2}-\frac{\pi}{2}+\ln2 =\ln2.$$ 6. Therefore, $$\int_0^1 \tan^{-1}(1-x+x^2)\,dx=\ln2.$$ So the correct option is **C**.

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Correct answer: C

Let $I=\int_0^1 \tan^{-1}(1-x+x^2)\,dx.$ We are given $2\int_0^1 \tan^{-1}x\,dx=\int_0^1 \cot^{-1}(1-x+x^2)\,dx.$
Use the standard identity for positive arguments: $\tan^{-1}t+\cot^{-1}t=\frac{\pi}{2}\quad (t>0).$ Here, $1-x+x^2=x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\quad \text{for }x\in[0,1].$ So, $\cot^{-1}(1-x+x^2)=\frac{\pi}{2}-\tan^{-1}(1-x+x^2).$ Hence,

=\frac{\pi}{2}-I.$$ 3. Therefore the given condition becomes $$2\int_0^1 \tan^{-1}x\,dx=\frac{\pi}{2}-I,$$ so $$I=\frac{\pi}{2}-2\int_0^1 \tan^{-1}x\,dx.$$ 4. Now evaluate $$\int_0^1 \tan^{-1}x\,dx.$$ Using integration by parts: Let $$u=\tan^{-1}x,\qquad dv=dx.$$ Then $$du=\frac{dx}{1+x^2},\qquad v=x.$$ Thus $$\int \tan^{-1}x\,dx=x\tan^{-1}x-\int \frac{x}{1+x^2}\,dx =x\tan^{-1}x-\frac12\ln(1+x^2).$$ So, $$\int_0^1 \tan^{-1}x\,dx=\left[x\tan^{-1}x-\frac12\ln(1+x^2)\right]_0^1 =\frac{\pi}{4}-\frac12\ln 2.$$ 5. Substitute into the expression for $I$: $$I=\frac{\pi}{2}-2\left(\frac{\pi}{4}-\frac12\ln2\right) =\frac{\pi}{2}-\frac{\pi}{2}+\ln2 =\ln2.$$ 6. Therefore, $$\int_0^1 \tan^{-1}(1-x+x^2)\,dx=\ln2.$$ So the correct option is **C**.