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

If ${u_n} = \int_0^{\pi /4} {{\tan ^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

Evaluate the definite integral: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}$.

$\int_{-a}^{a} \sin x \, f(\cos x) \, dx = $

$\int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx = $

The number of elements in the set $S = \{x : x \in [0, 100] \text{ and } \int_{0}^{x} t^{2} \sin(x-t) dt = x^{2}\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module