int frac{x^3}{sqrt{1+x^2}} dx is equal to | vedclass

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

Question

(C) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx$. We can rewrite the integral as $I = \int x^2 \cdot \frac{x}{\sqrt{1+x^2}} dx$. Using integration by pa

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx$. We can rewrite the integral as $I = \int x^2 \cdot \frac{x}{\sqrt{1+x^2}} dx$. Using integration by parts,let $u = x^2$ and $dv = \frac{x}{\sqrt{1+x^2}} dx$. Then $du = 2x dx$ and $v = \int (1+x^2)^{-1/2} x dx = \sqrt{1+x^2}$. Applying the formula $\int u dv = uv - \int v du$: $I = x^2 \sqrt{1+x^2} - \int 2x \sqrt{1+x^2} dx$. For the remaining integral,let $t = 1+x^2$,so $dt = 2x dx$. $I = x^2 \sqrt{1+x^2} - \int t^{1/2} dt = x^2 \sqrt{1+x^2} - \frac{t^{3/2}}{3/2} + C$. $I = x^2 \sqrt{1+x^2} - \frac{2}{3}(1+x^2)^{3/2} + C$.

$\int \frac{x^3}{\sqrt{1+x^2}} dx$ is equal to

Similar Questions

$\int x \cos x \, dx = $

If $\frac{d}{dx}f(x) = x\cos x + \sin x$ and $f(0) = 2$,then $f(x) = $

$\int (\log x)^3 x^5 dx = $

$\int \frac{x + \sin x}{1 + \cos x} \, dx$ is equal to

$\int (x^2 + 3x + 2) e^x dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module