Question

$\int \frac{x^2\left(x^4+4\right)}{x^2+4}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2\left(x^4+4\right)}{\left(x^2+4\right)}dx \\ =\int \left(\frac{x^6+4x^2}{x^2+4}\right)dx \\ \mathrm{Now}, \\ {x}^2+4\stackrel{x^4-4x^2+20}{\overline{)x^6+4x^2}} \\ {x}^6+4x^4 \\ \underline{--} \\ -4x^4+4x^2 \\ -4x^4-16x^2 \\ \underline{++} \\ 20x^2 \\ 20x^2+80 \\ \underline{--} \\ \underline{-80} \\ \mathrm{Therefore},\frac{x^2\left(x^4+4\right)}{\left(x^2+4\right)}=\left(x^4-4x^2+20\right)-\frac{80}{x^2+4} \\ I=\int \frac{x^2\left(x^4+4\right)}{\left(x^2+4\right)}dx \\ =\int \left(x^4-4x^2+20\right)dx-80\int \frac{{\displaystyle dx}}{x^2+2^2} \\ =\int x^{4}dx-4\int x^{2}dx+20\int dx-80\int \frac{dx}{x^2+2^2} \\ =\frac{x^{4+1}}{4+1}-4\left[\frac{x^3}{3}\right]+20\left(x\right)-80\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C \\ =\frac{x^5}{5}-\frac{4}{3}{x}^3+20x-40{\tan }^{-1}\left(\frac{x}{2}\right)+C \end{gathered}$$