int_0^infty {frac{{x,dx}}{{(1 + x)(1 + {x^2}) | 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 partial fractions,we write $\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx +

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(A) Let $I = \int_0^\infty \frac{x}{(1 + x)(1 + x^2)} dx$. Using partial fractions,we write $\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$. Solving for constants,we get $x = A(1 + x^2) + (Bx + C)(1 + x)$. For $x = -1$,$-1 = A(2) \implies A = -1/2$. Comparing coefficients of $x^2$,$0 = A + B \implies B = 1/2$. Comparing constants,$0 = A + C \implies C = 1/2$. Thus,$\frac{x}{(1 + x)(1 + x^2)} = \frac{1}{2} \left( \frac{x + 1}{1 + x^2} - \frac{1}{1 + x} ight) = \frac{1}{2} \left( \frac{x}{1 + x^2} + \frac{1}{1 + x^2} - \frac{1}{1 + x} ight)$. Integrating from $0$ to $\infty$: $I = \frac{1}{2} \left[ \frac{1}{2} \ln(1 + x^2) + 	an^{-1}(x) - \ln(1 + x) ight]_0^\infty$. $I = \frac{1}{2} \left[ \frac{1}{2} \ln\left(\frac{1 + x^2}{(1 + x)^2}ight) + 	an^{-1}(x) ight]_0^\infty$. As $x 	o \infty$,$\frac{1 + x^2}{(1 + x)^2} 	o 1$,so $\ln(1) = 0$. $I = \frac{1}{2} [0 + \frac{\pi}{2} - (0 + 0)] = \frac{\pi}{4}$.

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

Similar Questions

Evaluate the definite integral $\int_0^4 x[x] \, dx$,where $[x]$ denotes the greatest integer function not greater than $x$.

Evaluate the integral: $\int_0^2 [x] \, dx + \int_0^2 |x-1| \, dx$,where $[x]$ denotes the greatest integer function.

$\int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ equals

Let $[ t ]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_{1}^{2} |2x - [3x]| dx$ is

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module