int frac{dx}{sqrt{x^2 + a^2}} is equal to | vedclass

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

Question

(C) The integral $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is a standard integral formula.By using the substitution $x = a 	an 	heta$,we have $dx = a \sec^

Vedclass · Accepted Answer

$\log |x + \sqrt{x^2 + a^2}| + c$

Vedclass · Answer

(C) The integral $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is a standard integral formula.By using the substitution $x = a 	an 	heta$,we have $dx = a \sec^2 	heta \ d	heta$.Then,$\sqrt{x^2 + a^2} = \sqrt{a^2 	an^2 	heta + a^2} = a \sec 	heta$.The integral becomes $\int \frac{a \sec^2 	heta \ d	heta}{a \sec 	heta} = \int \sec 	heta \ d	heta$.The integral of $\sec 	heta$ is $\log |\sec 	heta + 	an 	heta| + c$.Substituting back $	an 	heta = \frac{x}{a}$ and $\sec 	heta = \frac{\sqrt{x^2 + a^2}}{a}$,we get $\log |\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}| + c = \log |\frac{x + \sqrt{x^2 + a^2}}{a}| + c = \log |x + \sqrt{x^2 + a^2}| - \log |a| + c$.Since $-\log |a| + c$ is a constant,we write the final result as $\log |x + \sqrt{x^2 + a^2}| + c$.

$\int \frac{dx}{\sqrt{x^2 + a^2}}$ is equal to

Similar Questions

Find an anti-derivative (or integral) of the function $\cos 3x$ by the method of inspection.

If $x \notin [2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}]$ and $n \in Z$,then $\int \sqrt{1 - \sin 2x} \, dx = $

$\int \frac{1}{x^2}(2 x+1)^3 d x$ is equal to

The integral $\int \sqrt{1 + 2\cot x(\csc x + \cot x)} \,dx$ for $0 < x < \frac{\pi}{2}$ is equal to (where $C$ is a constant of integration):

Integrate the function $e^{2x+3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module