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

$\int {\frac{{dx}}{{{x^2}{{(1 + {x^4})}^{\frac{3}{4}}}}}} $ is equal to- (where $C$ is constant of integration)

If $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$,then $\int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx = $

$\int \frac{dx}{(a^2 + x^2)^{3/2}}$ is equal to

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$

$\int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module