int_{-1}^{1} log left( frac{1+x}{1-x} right) | vedclass

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

Question

(C) Let $f(x) = \log \left( \frac{1+x}{1-x} \right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left( \frac{1+(-x)}{

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $f(x) = \log \left( \frac{1+x}{1-x} \right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left( \frac{1+(-x)}{1-(-x)} \right) = \log \left( \frac{1-x}{1+x} \right) = \log \left( \left( \frac{1+x}{1-x} \right)^{-1} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.Therefore,$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = 0$.

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

Similar Questions

$\int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx =$

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

If $I = \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$,then the value of $I$ is

$\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=$

The value of $\int_0^1 \tan^{-1}(1-x+x^2) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module