For 0

Question

(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 $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$.

Similar Questions

For $n \geq 2$,let $I_n = \int_0^{\pi / 4} \tan^n x \, dx$ and $F_n = I_n + I_{n-2}$. Then,$F_n - F_{n+1} =$

If $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$,then $f\left(\frac{2 \pi}{3}\right) =$

If $\int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = f(x) + C$ where $C$ is the constant of integration,then $f(\frac{\pi}{6}) - f(\frac{\pi}{4})$ is equal to

Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e(1 + 2\sqrt{2})$,then $\alpha^4$ is equal to $..........$.

If $\int \frac{5 \tan (x)}{\tan (x)-2} d x = x + a \log |\sin (x) - 2 \cos (x)| + k$,then $a$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module