Question

Evaluate the following integrals:
$\int \frac{x^3}{x^4+x^2+1}dx$

Answer 1

$$\begin{gathered} I=\int \frac{x^3}{x^4+x^2+1}dx \\ =\int \frac{x^2·x}{{\left(x^2\right)}^2+x^2+1}dx \\ \mathrm{Let}{x}^2=t\mathrm{or}2xdx=dt \\ \Rightarrow I=\frac{1}{2}\int \frac{t}{t^2+t+1}dt \\ =\frac{1}{4}\int \frac{2t}{t^2+t+1}dt \\ =\frac{1}{4}\int \frac{2t+1-1}{t^2+t+1}dt \end{gathered}$$

$$\begin{gathered} =\frac{1}{4}\int \left[\frac{\left(2t+1\right)}{\left(t^2+t+1\right)}-\frac{1}{\left(t^2+t+1\right)}\right]dt \\ =\frac{1}{4}\left[\log \left|t^2+t+1\right|-\int \frac{1}{\left(t^2+t+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}}\right)}dt\right] \\ =\frac{1}{4}\left[\log \left|t^2+t+1\right|-\int \frac{1}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dt\right] \\ =\frac{1}{4}\left[\log \left|t^2+t+1\right|-\frac{2}{\sqrt{3}}\tan \frac{\left(t+{\displaystyle \frac{1}{2}}\right)}{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}\right]+c \\ =\frac{1}{4}\left[\log \left|t^2+t+1\right|-\frac{2}{\sqrt{3}}\tan \left(\frac{2t+1}{{\displaystyle \sqrt{3}}}\right)\right]+c \end{gathered}$$

$=\frac{1}{4}\left[\log \left|x^4+x^2+1\right|-\frac{2}{\sqrt{3}}\tan \left(\frac{2x^2+1}{{\displaystyle \sqrt{3}}}\right)\right]+c$