int frac{x^3}{sqrt{x^2 + 2}} , dx = | vedclass

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

Question

(B) Let $I = \int \frac{x^3}{\sqrt{x^2 + 2}} \, dx$. Substitute $x^2 + 2 = t^2$,which implies $2x \, dx = 2t \, dt$ or $x \, dx = t \, dt$. Also,$x^2

Vedclass · Accepted Answer

$\frac{1}{3}(x^2 + 2)^{3/2} - 2(x^2 + 2)^{1/2} + c$

Vedclass · Answer

(B) Let $I = \int \frac{x^3}{\sqrt{x^2 + 2}} \, dx$. Substitute $x^2 + 2 = t^2$,which implies $2x \, dx = 2t \, dt$ or $x \, dx = t \, dt$. Also,$x^2 = t^2 - 2$. Substituting these into the integral: $I = \int \frac{(t^2 - 2)}{t} \cdot t \, dt = \int (t^2 - 2) \, dt$. Integrating with respect to $t$: $I = \frac{t^3}{3} - 2t + c$. Substituting $t = \sqrt{x^2 + 2}$ back into the expression: $I = \frac{1}{3}(x^2 + 2)^{3/2} - 2(x^2 + 2)^{1/2} + c$.

$\int \frac{x^3}{\sqrt{x^2 + 2}} \, dx = $

Similar Questions

The value of $\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ for $x>0$ is

$\int \sin^5 x \cos^4 x \, dx = $

$\int \frac{\ln |x|}{x\sqrt{1 + \ln |x|}} \, dx$ equals :

Let $\int \frac{x^{1/2}}{\sqrt{1-x^3}} dx = \frac{2}{3} g(f(x)) + c$; then

$\int \frac{x}{1+x^4} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module