int_{0}^{1} tan ^{-1}left(frac{2x-1}{1+x-x^{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{2x-1}{1+x-x^{2}}ight) dx$. We can rewrite the argument of the inverse tangent function as: $\frac{2x

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{2x-1}{1+x-x^{2}}ight) dx$. We can rewrite the argument of the inverse tangent function as: $\frac{2x-1}{1+x-x^{2}} = \frac{x - (1-x)}{1 + x(1-x)}$. Using the identity $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left(\frac{a-b}{1+ab}ight)$,we have: $I = \int_{0}^{1} (	an^{-1} x - 	an^{-1}(1-x)) dx$ ... $(1)$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{1} (	an^{-1}(1-x) - 	an^{-1}(1-(1-x))) dx = \int_{0}^{1} (	an^{-1}(1-x) - 	an^{-1} x) dx$ ... $(2)$. Adding equations $(1)$ and $(2)$: $2I = \int_{0}^{1} (	an^{-1} x - 	an^{-1}(1-x) + 	an^{-1}(1-x) - 	an^{-1} x) dx = \int_{0}^{1} 0 dx = 0$. Therefore,$I = 0$.

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

Similar Questions

The value of $\int_{-\pi/2}^{\pi/2} (3\sin x + \sin^3 x) \, dx$ is

For $n \in N$,the value of $\int_{0}^{n\pi + V} \sqrt{\frac{1 + \cos 2x}{2}} dx$ is . . . (where $\frac{\pi}{2} < V < \pi$)

For any integer $n$, the value of the integral $\int_0^\pi e^{\cos^2 x} \cos^3(2n+1)x \, dx$ is:

The integral $\int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$ is equal to

The value of $\int_1^3 \sqrt{3 + x^3} \,dx$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module