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

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(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

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

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

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