Evaluate the following integrals: x^5 1+x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}$

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Answer

$\text{I}=\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}\ ....(1)$ Let $1+\text{x}^3=\text{t}^2$ then, $\text{d}\big(1+\text{x}^3\big)=\text{d}\big(\text{t}^2\big)$ $\Rightarrow3\text{x}^2\text{dx}=\text{dt }2\text{t}$ $\Rightarrow\text{dx}=\frac{\text{dt}}{3\text{x}^2}\text{ 2}\text{t}$ Putting $1+\text{x}^3=\text{t}^2$ and $\text{dx}=\frac{2\text{t}}{3\text{x}^2}\text{ dt}$ in equation (1), we get,,$\text{I}=\int\frac{\text{x}^5}{\sqrt{{t}^2}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$
$=\int\frac{\text{x}^5}{\text{t}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$ $=\frac{2}{3}\int\text{x}^3\text{dt}$ $=\frac{2}{3}\int\big(\text{t}^2-1\big)\text{dt}$ $=\frac{2}{3}\times\frac{\text{t}^3}{3}-\frac{2}{3}\text{t}+\text{C}$ $\text{I}=\frac{2}{9}\big(1+\text{x}^3\big)^{\frac{3}{2}}-\frac{2}{3}\sqrt{1+\text{x}^3}+\text{C}$

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