Mathematics - Indefinite Integration Question with Solution | TestHub

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

Putting $1+2{\mathrm{x}}^2={\mathrm{t}}^2$

$4\mathrm{xdx}=2\mathrm{tdt}$

$\int \frac{\frac{{\mathrm{t}}^2-1}{2}\cdot \frac{\mathrm{t}}{2}\mathrm{dt}}{\mathrm{t}}\Rightarrow \frac{1}{4}\int \left({\mathrm{t}}^2-1\right)\mathrm{dt}$ $\Rightarrow \frac{1}{4}\left(\frac{{\mathrm{t}}^3}{3}-\mathrm{t}\right)+\mathrm{c}\Rightarrow \frac{1}{12}\mathrm{t}\left({\mathrm{t}}^2-3\right)+\mathrm{c}\Rightarrow \frac{1}{6}{\left(1+2{\mathrm{x}}^2\right)}^{1/2}\left({\mathrm{x}}^2-1\right)+\mathrm{c}$ $f(\mathrm{x})=\frac{1}{6}{\left(1+2{\mathrm{x}}^2\right)}^{1/2}\left({\mathrm{x}}^2-1\right)+\mathrm{c}$

which passes through $(1,2)$

$\therefore 2=0+c\therefore c=2$

$f(\mathrm{x})=\frac{1}{6}{\left(1+2{\mathrm{x}}^2\right)}^{1/2}\left({\mathrm{x}}^2-1\right)+2$

$\therefore \mathrm{m}+\mathrm{n}=6+2=8$