int_{0}^{infty} frac{dx}{(x^{2}+4)(x^{2}+9)} | vedclass

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

Question

(B) Let $I = \int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)}$.Using partial fractions,we write $\frac{1}{(x^{2}+4)(x^{2}+9)} = \frac{A}{x^{2}+4} + \fr

Vedclass · Accepted Answer

$\frac{\pi}{60}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)}$.Using partial fractions,we write $\frac{1}{(x^{2}+4)(x^{2}+9)} = \frac{A}{x^{2}+4} + \frac{B}{x^{2}+9}$.$1 = A(x^{2}+9) + B(x^{2}+4)$.Setting $x^{2} = -4$,we get $1 = A(5) \implies A = \frac{1}{5}$.Setting $x^{2} = -9$,we get $1 = B(-5) \implies B = -\frac{1}{5}$.So,$I = \frac{1}{5} \int_{0}^{\infty} \left( \frac{1}{x^{2}+4} - \frac{1}{x^{2}+9} ight) dx$.Using the formula $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \frac{1}{5} \left[ \frac{1}{2} 	an^{-1}(\frac{x}{2}) - \frac{1}{3} 	an^{-1}(\frac{x}{3}) ight]_{0}^{\infty}$.Evaluating the limits:$I = \frac{1}{5} \left[ (\frac{1}{2} \cdot \frac{\pi}{2} - 0) - (\frac{1}{3} \cdot \frac{\pi}{2} - 0) ight]$.$I = \frac{1}{5} \left( \frac{\pi}{4} - \frac{\pi}{6} ight) = \frac{1}{5} \left( \frac{3\pi - 2\pi}{12} ight) = \frac{1}{5} \cdot \frac{\pi}{12} = \frac{\pi}{60}$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

$\int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)} = $

Similar Questions

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x=$

$\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$ is equal to (where $p$ and $q$ are integers).

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

The value of $\int_{0}^{1} |3x^2 - 1| dx$ is

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:
$t \ (\text{in second})$ $0$ $2$ $4$ $6$ $8$ $10$
$v \ (\text{in m/s})$ $0$ $12$ $16$ $20$ $35$ $60$

Vedclass Test Series

Exam Paper Generator

Online Exam Module