Evaluate the definite integral: int_{0}^{inft | vedclass

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

Question

(A) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using the property $\int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(1/x) \cdot \f

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using the property $\int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(1/x) \cdot \frac{1}{x^2} \, dx$,we substitute $x = 1/t$,so $dx = -1/t^2 \, dt$. As $x 	o 0, t 	o \infty$ and as $x 	o \infty, t 	o 0$. $I = \int_{\infty}^{0} \frac{1/t}{(1 + 1/t)(1 + 1/t^2)} \cdot (-1/t^2) \, dt = \int_{0}^{\infty} \frac{1/t}{(\frac{t+1}{t})(\frac{t^2+1}{t^2})} \cdot \frac{1}{t^2} \, dt = \int_{0}^{\infty} \frac{1}{t(t+1)(t^2+1)/t^3} \cdot \frac{1}{t^2} \, dt = \int_{0}^{\infty} \frac{1}{(t+1)(t^2+1)} \, dt$. Now,$2I = \int_{0}^{\infty} \frac{x}{(1+x)(1+x^2)} \, dx + \int_{0}^{\infty} \frac{1}{(1+x)(1+x^2)} \, dx = \int_{0}^{\infty} \frac{x+1}{(1+x)(1+x^2)} \, dx = \int_{0}^{\infty} \frac{1}{1+x^2} \, dx$. $2I = [	an^{-1}(x)]_{0}^{\infty} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. Therefore,$I = \frac{\pi}{4}$.

Evaluate the definite integral: $\int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$

Similar Questions

Let ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$ and ${I_2} = \int_a^{\pi - a} {f(\sin x)dx}$,then ${I_2}$ is equal to

Let $m, n, p, q$ be four positive integers. If $\int_0^{2 \pi} \sin^m x \cos^n x \, dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x \, dx$, $\int_0^{2 \pi} \sin^p x \cos^n x \, dx = 0$, $\int_0^{\pi} \sin^p x \cos^q x \, dx = 0$, $a = m + n + p$ and $b = m + n + q$, then:

On the real line $R$,we define two functions $f$ and $g$ as follows:
$f(x) = \min \{x - [x], 1 - x + [x]\}$
$g(x) = \max \{x - [x], 1 - x + [x]\}$
where $[x]$ denotes the greatest integer function. The positive integer $n$ for which $\int_0^n (g(x) - f(x)) \, dx = 100$ is:

$\int_{-5}^{5} \left[ \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} \right] dx = $

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module