The value of the integral int_0^infty frac{lo | vedclass

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

Question

(B) Let $I = \int_0^\infty \frac{\log_e(x)}{x^2+4} dx$.Substitute $x = 2t$,so $dx = 2 dt$. As $x 	o 0, t 	o 0$ and as $x 	o \infty, t 	o \infty$.$

Vedclass · Accepted Answer

$\frac{\pi \log_e(2)}{4}$

Vedclass · Answer

(B) Let $I = \int_0^\infty \frac{\log_e(x)}{x^2+4} dx$.Substitute $x = 2t$,so $dx = 2 dt$. As $x 	o 0, t 	o 0$ and as $x 	o \infty, t 	o \infty$.$I = \int_0^\infty \frac{\log_e(2t)}{4t^2+4} (2 dt) = \frac{1}{2} \int_0^\infty \frac{\log_e(2) + \log_e(t)}{t^2+1} dt$.$I = \frac{\log_e(2)}{2} \int_0^\infty \frac{1}{t^2+1} dt + \frac{1}{2} \int_0^\infty \frac{\log_e(t)}{t^2+1} dt$.For the second integral,let $t = 1/u$,then $dt = -1/u^2 du$. The limits change from $\infty$ to $0$.$\int_0^\infty \frac{\log_e(t)}{t^2+1} dt = \int_\infty^0 \frac{\log_e(1/u)}{(1/u)^2+1} (-1/u^2) du = \int_0^\infty \frac{-\log_e(u)}{1+u^2} du = -\int_0^\infty \frac{\log_e(u)}{u^2+1} du$.This implies $\int_0^\infty \frac{\log_e(t)}{t^2+1} dt = 0$.Thus,$I = \frac{\log_e(2)}{2} [\arctan(t)]_0^\infty = \frac{\log_e(2)}{2} (\frac{\pi}{2} - 0) = \frac{\pi \log_e(2)}{4}$.

The value of the integral $\int_0^\infty \frac{\log_e(x)}{x^2+4} dx$ is:

Similar Questions

Evaluate the definite integral: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

Let the domain of the function $f(x) = \log_{4}(\log_{5}(\log_{3}(18x - x^{2} - 77)))$ be $(a, b)$. Then the value of the integral $\int_{a}^{b} \frac{\sin^{3} x}{\sin^{3} x + \sin^{3}(a + b - x)} dx$ is equal to $.....$

The value of $\int_{0}^{\pi /2} \frac{\sin^{2/3} x}{\sin^{2/3} x + \cos^{2/3} x} dx$ is

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x=$

$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module