int frac{d x}{sqrt{x}(x+9)} is equal to | vedclass

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

Question

(B) Let $I = \int \frac{d x}{\sqrt{x}(x+9)}$.Substitute $x = t^2$, which implies $d x = 2t \, dt$.Substituting these into the integral:$I = \int \frac

Vedclass · Accepted Answer

$\frac{2}{3} 	an ^{-1}\left(\frac{\sqrt{x}}{3}ight)+C$

Vedclass · Answer

(B) Let $I = \int \frac{d x}{\sqrt{x}(x+9)}$.Substitute $x = t^2$, which implies $d x = 2t \, dt$.Substituting these into the integral:$I = \int \frac{2t \, dt}{t(t^2 + 9)} = \int \frac{2 \, dt}{t^2 + 3^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = 2 \cdot \frac{1}{3} 	an^{-1}(\frac{t}{3}) + C$.Substituting $t = \sqrt{x}$ back into the expression:$I = \frac{2}{3} 	an^{-1}(\frac{\sqrt{x}}{3}) + C$.

$\int \frac{d x}{\sqrt{x}(x+9)}$ is equal to

Similar Questions

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

Find the following integral: $\int \frac{\sin x}{\sin (x+a)} d x$

Integrate the following function with respect to $x:$
$\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}$

$\int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} = $

If $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$,then $e^{f(x)}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module