Evaluation of $\int\frac{1}{(x^3+1)^2}\mathrm dx \cdots$

Question

$\displaystyle \int\frac{1}{(x^3+1)^2} \mathrm dx$

$\bf{My\; Try}::$ Using Integration by parts

Let $\displaystyle I = \int\frac{1}{(x^3+1)}\cdot 1\; dx = \frac{1}{(x^3+1)}\cdot x + \int\frac{3x^2\cdot x}{(x^3+1)^2}dx$

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

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

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

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

Now Help Required

Thanks

Your partial fraction expansion is not correct, but your idea to do this is spot on. — Amzoti, Nov 09 '13 at 09:32
Related : http://math.stackexchange.com/questions/556115/compute-int-0-infty-fracdx1x3 — lab bhattacharjee, Nov 09 '13 at 09:36

score 2 · Accepted Answer · answered Nov 09 '13 at 10:04

Good Work so far.

$$ \begin{equation} \begin{split} \int\frac{1}{x^3+1} \mathrm{d}x & = \int\frac{1}{3(x+1)} \mathrm{d}x + \int\frac{2-x}{3(x^2-x+1)} \mathrm{d}x \\ & = \frac{1}{3} \ln\left|x+1\right| - \frac{1}{6} \int \frac{2x-1 -3}{x^2-x+1}\mathrm{d}x \\ & = \frac{1}{3} \ln\left|x+1\right| - \frac{1}{6} \int\frac{2x-1}{x^2-x+1}\mathrm{d}x + \frac{1}{2} \int\frac{1}{x^2-x+1}\mathrm{d}x \\ & = \frac{1}{3} \ln\left|x+1\right| - \frac{1}{6} \ln\left|x^2-x+1\right| + \frac{1}{2} \int\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\mathrm{d}x \\ & = \frac{1}{3} \ln\left|x+1\right| - \frac{1}{6} \ln\left|x^2-x+1\right| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right) \\ \end{split} \end{equation} $$

Evaluation of $\int\frac{1}{(x^3+1)^2}\mathrm dx \cdots$

1 Answers1