# integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$

What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$

I know about partial fractions: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$

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

$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3+Gx^2+H+I}{(x^2-x+1)^2}\right)dx$$ and after this solving is easy, i was trying to do the same many times, but i can't find coefficients because mistakes or something other.

I want to know about another methods to solve it.

• Your partial fractions are wrong: the second one, numerator is constant, and fourth one, numerator is linear. – samjoe Mar 19 '17 at 7:30
• @samjoe oh, exactly, you are right, thank you very much – Shmuser Mar 19 '17 at 7:36
• – user236182 Sep 3 '17 at 19:45

HINT:

First integrate by parts,

$$3I=\int\dfrac1x\cdot\dfrac{3x^2}{(1+x^3)^2}dx=-\dfrac1{x(1+x^3)}+\int\dfrac{dx}{x^2(1+x^3)}$$

Now use Partial fraction $$\dfrac1{x^2(1+x^3)}=\dfrac Ax+\dfrac B{x^2}+\dfrac C{x+1}+\dfrac{Dx+E}{x^2-x+1}$$

• It should be $-\int \frac{dx}{x^2\left(1+x^3\right)}$. – user236182 Aug 17 '17 at 19:11

If you can use the Internet, WolframAlpha (link) gives

$$\frac{x}{(x^3+1)^2}=\frac{x-1}{9(x^2-x+1)}+$$

$$+\frac{1}{3(x^2-x+1)^2}-\frac{1}{9(x+1)}-$$

$$-\frac{1}{9(x+1)^2}$$

The most difficult part, if you can't use non-real numbers ($x^2-x+1$ can't be non-trivially factored over the real numbers, partial fractions with real numbers can't be used), could be integrating $\int \frac{1}{(x^2-x+1)^2}\, dx$. See the last part of this answer for that.

$$\int \frac{1}{x+1}\, dx=\int \frac{1}{x+1}\, d(x+1)$$

$$=\ln|x+1|+C$$

$$\int \frac{1}{(x+1)^2}\, dx=\int \frac{1}{(x+1)^2}\, d(x+1)$$

$$=-\frac{1}{x+1}+C$$

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

Notice that $(x^2-x+1)'=2x-1$.

$$=\frac{1}{2}\int \frac{d\left(x^2-x+1\right)}{x^2-x+1}-\frac{1}{2}\int \frac{1}{x^2-x+1}\, dx$$

$$=\frac{1}{2}\ln |x^2-x+1|-\frac{1}{2}\int \frac{d\left(x-\frac{1}{2}\right)}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}$$

$$=\frac{1}{2}\ln|x^2-x+1|-\frac{1}{2}\cdot \frac{1}{\frac{\sqrt{3}}{2}}\cdot \arctan\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$$

You can find $\int \frac{1}{(x^2-x+1)^2}\, dx$ without using non-real numbers, we'll start by finding $\int \frac{1}{x^2-x+1}\, dx$ using integration by parts.

$$\int u\, dv=uv-\int v\, du$$

$$u=\frac{1}{x^2-x+1}$$

$$du=\frac{-(2x-1)}{(x^2-x+1)^2}\, dx$$

$$dv=dx,\, v=x$$

$$\int \frac{1}{x^2-x+1}\, dx=\frac{x}{x^2-x+1}+$$

$$+\int \frac{x(2x-1)}{(x^2-x+1)^2}\, dx$$

If you can use the Internet, WolframAlpha (link) gives

$$\frac{x(2x-1)}{\left(x^2-x+1\right)^2}=\frac{x-2}{\left(x^2-x+1\right)^2}+$$

$$+\frac{2}{x^2-x+1}$$

I've already found $\int \frac{1}{x^2-x+1}\, dx$ in this answer before.

$$\int \frac{x-2}{\left(x^2-x+1\right)^2}=$$

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

Again, notice that $(x^2-x+1)'=2x-1$.

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

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

Now find $\int \frac{1}{(x^2-x+1)^2}\, dx$ in terms of $\int \frac{1}{x^2-x+1}\, dx$, which I've already found in this answer before.

More generally, $\int \frac{1}{\left(x^2+bx+c\right)^n}\, dx$, where $n\ge 2$, $n\in\mathbb Z$, $b^2-4c<0$, $b,c\in\mathbb R$ could be integrated using real numbers (notice that $x^2+bx+c$ has only non-real roots, only non-real non-trivial factorization, so partial fractions with real numbers can't be used). See here[1] and here[2] for formulas for $\int \frac{1}{(x^2+bx+c)^n}\, dx$, $\int \frac{x}{(x^2+bx+c)^n}\, dx$.

In particular, $[2]$ link has an example of how to integrate $\int \frac{1}{\left(x^2+1\right)^2}\, dx$. It's similar to what I've done here.

More generally, $$\left(\frac{1}{\left(x^2+bx+c\right)^n}\right)'=\frac{-(2x+b)n}{\left(x^2+bx+c\right)^{n+1}}$$

$$\left(\frac{x}{\left(x^2+bx+c\right)^n}\right)'=\frac{x^2+bx+c-nx(2x+b)}{\left(x^2+bx+c\right)^{n+1}}$$

I hope I haven't made any mistakes. This is a long answer.

\begin{align} \int\frac{x\,\mathrm{d}x}{\left(x^3+1\right)^2} &=-\frac13\int\frac1x\,\mathrm{d}\frac1{x^3+1}\tag{1}\\ &=-\frac1{3x\left(x^3+1\right)}-\frac13\int\frac{\mathrm{d}x}{x^2\left(x^3+1\right)}\tag{2}\\ &=-\frac1{3x\left(x^3+1\right)}-\frac13\int\left(\frac1{x^2}-\frac{x}{x^3+1}\right)\,\mathrm{d}x\tag{3}\\ &=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}+\frac19\int\left(\frac{x+1}{x^2-x+1}-\frac1{x+1}\right)\,\mathrm{d}x\tag{4}\\ &=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}-\frac19\log(x+1)+\frac19\int\frac{x-\frac12+\frac32}{\left(x-\frac12\right)^2+\frac34}\,\mathrm{d}x\tag{5}\\ &=\small\frac{x^2}{3\left(x^3+1\right)}-\frac19\log(x+1)+\frac1{18}\log\left(x^2-x+1\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{6}\\ &=\frac{x^2}{3\left(x^3+1\right)}+\frac1{18}\log\left(\frac{x^3+1}{(x+1)^3}\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{7} \end{align} Explanation:
$(1)$: prepare to integrate by parts
$(2)$: integrate by parts
$(3)$: partial fractions
$(4)$: partial fractions
$(5)$: integrate
$(6)$: integrate
$(7)$: simplify