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

(C) 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} + \f

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(C) 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 $A = -1/2$,$B = 1/2$,and $C = 1/2$. Thus,$I = \int_{0}^{\infty} \left( \frac{-1/2}{1 + x} + \frac{1/2 x + 1/2}{1 + x^2} ight) \, dx$. $I = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{1 + x^2} \, dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x^2}$. Evaluating the integral: $I = \left[ -\frac{1}{2} \ln(1 + x) + \frac{1}{4} \ln(1 + x^2) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$. $I = \left[ \frac{1}{4} \ln\left( \frac{1 + x^2}{(1 + x)^2} ight) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$. As $x 	o \infty$,$\frac{1 + x^2}{(1 + x)^2} 	o 1$,so $\ln(1) = 0$. $I = (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (0 + 0) = \frac{\pi}{4}$.

$\int_{0}^{\infty} \frac{x \, dx}{(1 + x)(1 + x^2)} = $

Similar Questions

Consider the equation $\int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$,where $a \in (-\infty, 0) \cup (1, \infty)$. Which of the following statements is/are $TRUE$?

$\int_0^1 \log (x+1) \, dx =$

The value of $\int_{a}^{b} \frac{x}{|x|} dx$,where $a < b < 0$,is

$\int_0^{1.5} x[x^2] dx = $

The value of $\alpha$ for which $4 \alpha \int_{-1}^{2} e^{-\alpha |x|} dx = 5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module