The value of int_{0}^{infty} frac{x}{(1+x)(x^ | vedclass

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

Question

(B) Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$.Using partial fractions,we write $\frac{x}{(1+x)(x^{2}+1)} = \frac{A}{1+x} + \frac{Bx+C}{x

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$.Using partial fractions,we write $\frac{x}{(1+x)(x^{2}+1)} = \frac{A}{1+x} + \frac{Bx+C}{x^{2}+1}$.Equating numerators: $x = A(x^{2}+1) + (Bx+C)(1+x) = (A+B)x^{2} + (B+C)x + (A+C)$.Comparing coefficients: $A+B=0$,$B+C=1$,$A+C=0$.Solving these,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.Thus,$I = \int_{0}^{\infty} \left( -\frac{1}{2(1+x)} + \frac{x+1}{2(x^{2}+1)} ight) dx = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1+x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{x^{2}+1} dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2}+1}$.Evaluating the integrals: $I = \left[ -\frac{1}{2} \ln(1+x) + \frac{1}{4} \ln(x^{2}+1) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$.Combining the logarithmic terms: $I = \left[ \frac{1}{4} \ln \left( \frac{(x^{2}+1)}{(1+x)^{2}} ight) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$.As $x 	o \infty$,$\frac{x^{2}+1}{(1+x)^{2}} 	o 1$,so $\ln(1) = 0$.At $x=0$,$\frac{1}{4} \ln(1) + \frac{1}{2} 	an^{-1}(0) = 0$.Therefore,$I = 0 + \frac{1}{2} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{4}$.

The value of $\int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$ is

Similar Questions

The value of the definite integral $\int_{0}^{1} e^{e^x}(1 + x e^x) dx$ is equal to

The number of positive solutions of the equation $\int_{0}^{x} (t - \{t\})^2 dt = 2(x - 1)$,where $\{ \}$ denotes the fractional part function,is:

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

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

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$,then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module