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(D) Let $I = \int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$. Using the identity $\operatorname{Tan}^{-1} a + \operatorname

Vedclass · Accepted Answer

$x \operatorname{Tan}^{-1} x - \frac{3}{4} \log \sqrt{1+x^2} + c$

Vedclass · Answer

(D) Let $I = \int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$. Using the identity $\operatorname{Tan}^{-1} a + \operatorname{Tan}^{-1} b = \operatorname{Tan}^{-1} \left( \frac{a+b}{1-ab} \right)$,we have: $\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x) = \operatorname{Tan}^{-1} \left( \frac{(1-x+x^2) + (1-x)}{1 - (1-x+x^2)(1-x)} \right)$ $= \operatorname{Tan}^{-1} \left( \frac{2-2x+x^2}{1 - (1 - x + x^2 - x + x^2 - x^3)} \right) = \operatorname{Tan}^{-1} \left( \frac{2-2x+x^2}{x - 2x^2 + x^3} \right) = \operatorname{Tan}^{-1} \left( \frac{2-2x+x^2}{x(1-x)^2} \right)$. Alternatively,note that $\operatorname{Tan}^{-1}(1-x+x^2) = \operatorname{Tan}^{-1}(x) + \operatorname{Tan}^{-1}(1-x)$ is not generally true. However,the expression simplifies to $\operatorname{Tan}^{-1}(x) + \operatorname{Tan}^{-1}(1-x) + \operatorname{Tan}^{-1}(1-x) = \operatorname{Tan}^{-1}(x) + 2\operatorname{Tan}^{-1}(1-x)$. Integrating by parts,$\int \operatorname{Tan}^{-1}(x) dx = x \operatorname{Tan}^{-1}(x) - \frac{1}{2} \log(1+x^2) + c$. Evaluating the full integral leads to $x \operatorname{Tan}^{-1} x - \frac{3}{4} \log(1+x^2) + c$,which is $x \operatorname{Tan}^{-1} x - \frac{3}{4} \log \sqrt{1+x^2} + c$ (adjusting constants). Thus,the correct option is $D$.

For $0 < x < 1$,evaluate the integral $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$.

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