intlimits_0^{frac{1}{2}} frac{1}{1 - x^2} ln | vedclass

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

Question

(A) Let $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$. Substitute $t = \ln \left( \frac{1 + x}{1 - x}

Vedclass · Accepted Answer

$\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$

Vedclass · Answer

(A) Let $I = \int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$. Substitute $t = \ln \left( \frac{1 + x}{1 - x} \right) = \ln(1 + x) - \ln(1 - x)$. Then $dt = \left( \frac{1}{1 + x} - \frac{-1}{1 - x} \right) dx = \left( \frac{1 - x + 1 + x}{1 - x^2} \right) dx = \frac{2}{1 - x^2} dx$. Thus,$\frac{dx}{1 - x^2} = \frac{1}{2} dt$. When $x = 0$,$t = \ln(1) = 0$. When $x = \frac{1}{2}$,$t = \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \ln \left( \frac{3/2}{1/2} \right) = \ln 3$. Therefore,$I = \int\limits_0^{\ln 3} t \cdot \frac{1}{2} dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_0^{\ln 3} = \frac{1}{4} \ln^2 3$. Since $\ln^2(1/3) = (\ln 1 - \ln 3)^2 = (-\ln 3)^2 = \ln^2 3$,the result is $\frac{1}{4} \ln^2 3$ or $\frac{1}{4} \ln^2 \left( \frac{1}{3} \right)$.

$\int\limits_0^{\frac{1}{2}} \frac{1}{1 - x^2} \ln \left( \frac{1 + x}{1 - x} \right) dx$ is equal to :

Similar Questions

$\int_0^a {{x^2}\sin {x^3}\,dx} $ equals

Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$.

Evaluate the definite integral $\int_{0}^{1} x e^{x^{2}} d x$.

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.

$\int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module