Question

Evaluate: $\int \frac{x}{1+x^3}dx$

Answer 1

$I=\int \frac{x}{1+x^3}dx$

Using partial fraction :- $\frac{x}{1+x^3}=\frac{x+1}{3(x^2-x+1)}-\frac{1}{3(x+1)}$

$\therefore I=\int \frac{x}{1+x^3}dx=\int \frac{x+1}{3(x^2-x+1)}-\frac{1}{3(x+1)}dx$

Now $\int \frac{x+1}{3(x^2-x+1)}dx=\frac{1}{3}\int \frac{x+1}{x^2-x+1}dx$

$=\frac{1}{3}\int \frac{x+1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}dx$

Now let $u=x-\frac{1}{2}$

$\therefore I=\frac{1}{3}\int \frac{2(2u+3)}{4u^2+3}du$

$=\frac{2}{3}[\int \frac{2u}{4u^2+3}du+\int \frac{3}{4u^2+3}du]$

$=\frac{2}{3}[\frac{1}{4}log|4u^2+3|+\frac{\sqrt{3}}{2}ta{n}^{-1}(\frac{2}{\sqrt{3}}u\left)\right]$

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

Now $\int \frac{1}{3(x+1)}dx=\frac{1}{3}log|x+1|$

$\therefore I=\frac{1}{6}(log|4x^2-4x+4|+2\sqrt{3}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right))-\left(\frac{1}{3}log|x+1|\right)+c$