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

If $\int x^3(\log x)^2 d x = x^4[A(\log x)^2 + B(\log x) + C] + K$, then find the value of $A + B + C$.

If $u = \int e^{ax} \cos bx \, dx$ and $v = \int e^{ax} \sin bx \, dx$,then $(a^2 + b^2)(u^2 + v^2) = $

If $I_n = \int (\log x)^n \, dx$,then $I_n + n I_{n-1} =$

$\int \log x^2 \, dx =$ . . . . . . $+ C$.

$\int e^x \sin x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module