Question
Integrate the functions in Exercises:
$\text{x}\sqrt{1+2\text{x}^2}$

Answer

$\text{Let}\text{ I}=\int\text{x}\sqrt{1+2\text{x}^2}\text{ dx} =\frac{1}{4}\int\sqrt{1+2\text{x}^2}(4\text{x dx}) \ \ \ \ \ \ ....\text{(i)} $
Putting $ 1+2\text{x}^2=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ 4\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow \ \ \ \ 4\text{x}\text{ dx}=\text{dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{4}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{4}\int\text{t}^{\frac{1}{2}}\text{ dt}$
$=\frac{1}{4}\frac{\text{t}^{^3/_2}}{{^3/_2}}+\text{c}=\frac{1}{4}\dot\ \frac{2}{3}\text{t}^{^3/_2 } +\text{c} $
$=\frac{1}{6}\big(1 + 2\text{x}^2\big)^{^3/_2}+\text{c} $

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