int_0^infty frac{dx}{(x + sqrt{x^2 + 1})^3} = | vedclass

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

Question

(A) Let $I = \int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3}$.Substitute $x = 	an 	heta$,so $dx = \sec^2 	heta \, d	heta$.When $x = 0$,$	heta = 0

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(A) Let $I = \int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3}$.Substitute $x = 	an 	heta$,so $dx = \sec^2 	heta \, d	heta$.When $x = 0$,$	heta = 0$. When $x 	o \infty$,$	heta 	o \frac{\pi}{2}$.The integral becomes:$I = \int_0^{\pi/2} \frac{\sec^2 	heta \, d	heta}{(	an 	heta + \sec 	heta)^3}$$I = \int_0^{\pi/2} \frac{\sec^2 	heta \, d	heta}{(\frac{\sin 	heta + 1}{\cos 	heta})^3} = \int_0^{\pi/2} \frac{\sec^2 	heta \cos^3 	heta}{(1 + \sin 	heta)^3} \, d	heta$$I = \int_0^{\pi/2} \frac{\cos 	heta}{(1 + \sin 	heta)^3} \, d	heta$Let $u = 1 + \sin 	heta$,then $du = \cos 	heta \, d	heta$.When $	heta = 0, u = 1$. When $	heta = \frac{\pi}{2}, u = 2$.$I = \int_1^2 u^{-3} \, du = \left[ \frac{u^{-2}}{-2} ight]_1^2 = \left[ -\frac{1}{2u^2} ight]_1^2$$I = -\frac{1}{2(4)} - (-\frac{1}{2(1)}) = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}$.

$\int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3} = $

Similar Questions

Let $0 < \alpha < \beta < 1$. Then $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x}$ is

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

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) d x=$, where $[.]$ denotes the Greatest Integer Function.

$\int_{0}^{5} \frac{d x}{x^{2}+2 x+10} = $

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module