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

If $\int \frac{1+\cos (4 x)}{\cot (x)-\tan (x)} d x=k \cos (4 x)+c$,then

$\int {(\sin^4 x - \cos^4 x) \, dx} = $

Let $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$ and $f(1)=1$. Then $f(5)=$

Find the following integral: $\int \left(x^{\frac{2}{3}} + 1\right) dx$

If $u$ and $v$ are functions of $x$, then $\int \frac{1}{v^3} (uv \frac{du}{dx} - u^2 \frac{dv}{dx}) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module