I would like some help with this problem:
Evaluate
$\displaystyle \int \frac{x}{x^3 + 1} dx$
There's always partial fraction decomposition. Can't think of a shorter way atm .. maybe Krizalid can think of something!
:
$\displaystyle \int \frac{x}{x^3 + 1} \ dx$
$\displaystyle = \int \frac{x}{(x+1)(x^2-x+1)} \ dx$
$\displaystyle = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x+1}{x^2-x+1}\right) dx$ (By partial fraction decomposition)
$\displaystyle = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x+1 {\color{red} \ + 1 - 1}}{x^2-x+1}\right) dx$
$\displaystyle = \int \left(-\frac{1}{3(x+1)} \ + \ \frac{1}{3} \cdot \frac{x-1}{x^2-x+1} \ + \ \frac{2}{3} \frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}}\right) \ dx$
$\displaystyle = \hdots$
Sure:
$\displaystyle \frac{1}{3}\left(\frac{x+1 {\color{red} \ - 1 + 1}}{x^2-x-1}\right)$
$\displaystyle =\frac{1}{3}\left(\frac{x-1}{x^2 - x + 1} + \frac{2}{x^2 - x - 1}\right)$
Completing the square in the bottom of the 2nd fraction:
$\displaystyle =\frac{1}{3}\left(\frac{x-1}{x^2 -x + 1} + \frac{2}{\left(x^2 - x {\color{red}\ + \left(\frac{1}{2}\right)^2} \right) + 1 {\color{red}- \left(\frac{1}{2}\right)^2}}\right)$
$\displaystyle = \frac{1}{3} \left(\frac{x-1}{x^2 - x + 1} + \frac{2}{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}}\right)$
Now multiply the $\displaystyle \frac{1}{3}$ in and you'll get what I had earlier. Use a simple sub for the first one and a trig sub for the other.