Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$

Question

I need to solve the following integrals:

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

and

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

My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?

Answer 1

The standard method uses partial fractions decomposition.

Here is how it begins for the first integral: factorout the denominator into irreducible factors with high-school identities: $$x^3+1=(x+1)(x^2-x+1)$$ Newt proceed to decompose into partial fractions: $$\frac{1}{(x^3+1)^2}=\frac{1}{(x+1)^2(x^2-x+1)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac {Cx+D}{x^2-x+1}+\frac{Ex+F}{(x^2-x+1)^2}.$$

Answer 2

$$I=\int{\frac{1}{(x^3+1)^2}}dx$$ $$I=-\int \frac {1}{3x^2}\color {red}{{\frac{-3x^2}{(x^3+1)^2}}}dx$$ The function in red is a derivative. $$I=-\int \frac {1}{3x^2}\color {red}{ \left ({\frac{1}{x^3+1}} \right )'}dx$$

The integral is now of the form: $$ \color {blue}{I=\int f(x) g'(x)dx}$$ Integrate by part $$ \color {blue}{\int f(x) g'(x)dx=f(x)g(x)-\int f'(x)g(x) dx}$$ $$I=- \frac {1}{3x^2}{\frac{1}{(x^3+1)}}-\frac {2}{3}\int \frac {1}{x^3}{\frac{1}{(x^3+1)}}dx$$ $$I_2=\int \frac {1}{x^3}{\frac{1}{(x^3+1)}}dx$$ $$I_2=\int \frac {dx}{x^3}-\int {\frac{dx}{(x^3+1)}}$$ $$I_2=- \frac {1}{2x^2}-\int {\frac{1}{(x^3+1)}}dx$$ So that we have : $$I=\frac x{3(x^3+1)}+\frac23\int\frac1{x^3+1}dx$$ Then use fraction decomposition method for that integral.

Answer 3

There's been a lot of this lately so what if we want to do $$\begin{align}\int\frac{dx}{(x^n+a)^m}&=\frac1a\int\frac{(x^n+a-x^n)dx}{(x^n+a)^m}\\ &=\frac1a\int\frac{dx}{(x^n+a)^{m-1}}-\frac1a\int\frac{x\cdot x^{n-1}dx}{(x^n+a)^m}\\ &=\frac1a\int\frac{dx}{(x^n+a)^{m-1}}+\frac x{a(m-1)n(x^n+a)^{m-1}}\\ &\quad-\frac1{a(m-1)n}\int\frac{dx}{(x^n+a)^{m-1}}\\ &=\frac{mn-n-1}{a(m-1)n}\int\frac{dx}{(x^n+a)^{m-1}}+\frac x{a(m-1)n(x^n+a)^{m-1}}\\ &=\frac{a^{m-1}\Gamma\left(m-\frac1n\right)\Gamma(m-1)}{a^m\Gamma\left(m-1-\frac1n\right)\Gamma(m)}\int\frac{dx}{(x^n+a)^{m-1}}+\frac x{a(m-1)n(x^n+a)^{m-1}}\end{align}$$ So we can say $$\begin{align}\frac{a^m\Gamma(m)}{\Gamma\left(m-\frac1n\right)}\int\frac{dx}{(x^n+a)^m}&=\frac{a^{m-1}\Gamma(m-1)}{\Gamma\left(m-1-\frac1n\right)}\int\frac{dx}{(x^n+a)^{m-1}}\\ &\quad+\frac{a^{m-1}\Gamma(m-1)x}{n\Gamma\left(m-\frac1n\right)(x^n+a)^{m-1}}\\ &=\frac a{\Gamma\left(1-\frac1n\right)}\int\frac{dx}{x^n+a}+\sum_{k=2}^m\frac{a^{k-1}\Gamma(k-1)x}{n\Gamma\left(k-\frac1n\right)(x^n+a)^{k-1}}\end{align}$$ For the purposes of this question, let $a=\pm1$, $\omega^3=-a$, then $$\frac1{x^3+a}=\sum_{k=1}^3\frac{A_k}{x-\omega^{2k-1}}$$ Where we can compute $$A_k=\lim_{x\rightarrow\omega^{2k-1}}\frac{x-\omega^{2k-1}}{x^3+a}=\frac1{3\omega^{4k-2}}=\frac{\omega^{2k-1}}{3(-a)^{2k-1}}=-\frac{\omega^{2k-1}}{3a}$$ Then $$\int\frac{dx}{x^3+a}=-\frac1{3a}\sum_{k=1}^3\omega^{2k-1}\int\frac{dx}{x-\omega^{2k-1}}=-\frac1{3a}\sum_{k=1}^3\omega^{2k-1}\ln(x-\omega^{2k-1})+C$$ So the integral is done, but we have a little cleaning up to do. Since $\omega^3=-a$, the other $2$ roots are complex conjugates: $\omega^5=\omega^*$ and we can write $$\begin{align}\omega\ln(x-\omega)+\omega^*\ln(x-\omega^*)&=\frac12(\omega+\omega^*)\ln(x-\omega)+\frac12(\omega-\omega^*)\ln(x-\omega)\\ &\quad+\frac12(\omega^*+\omega)\ln(x-\omega^*)+\frac12(\omega^*-\omega)\ln(x-\omega^*)\\ &=\Re\omega\ln(x^2-2\Re\omega x+1)+2i\Im\omega\cdot i\tan^{-1}\left(\frac{-\Im\omega}{x-\Re\omega}\right)\end{align}$$ For $a=1$, $\omega=e^{\pi i/3}=\cos(\pi/3)+i\sin(\pi/3)=\frac12+\frac i2\sqrt3$, while for $a=-1$, $\omega=e^{2\pi i/3}=-\frac12+\frac i2\sqrt3=\frac a2+\frac i2\sqrt3$ in both cases. Thus our formula becomes $$\int\frac{dx}{x^3+a}=-\frac1{3a}\left\{\frac a2\ln(x^2-ax+1)+\sqrt3\tan^{-1}\left(\frac{\sqrt3}{2x-a}\right)-a\ln(x+a)\right\}$$ Thus for $n=3$, $m=2$, and $a=1$ we get $$\begin{align}\int\frac{dx}{(x^3+1)^2}&=\frac{0!x}{(1)1!(3)(x^3+1)}+\frac{\left(1-\frac13\right)}{(1)1!}\left(-\frac1{3(1)}\right)\left\{\frac{(1)}2\ln(x^2-(1)x+1)\right.\\ &\quad\left.+\sqrt3\tan^{-1}\left(\frac{\sqrt3}{2x-(1)}\right)-(1)\ln(x+(1))\right\}+C\\ &=\frac{x}{3(x^3+1)}-\frac29\left\{\frac12\ln(x^2-x+1)\right.\\ &\quad\left.+\sqrt3\tan^{-1}\left(\frac{\sqrt3}{2x-1}\right)-\ln(x+1)\right\}+C\end{align}$$ Checked while for $n=3$, $m=2$, and $a=-1$ we have $$\begin{align}\int\frac{dx}{(x^3-1)^2}&=\frac{0!x}{(-1)1!(3)(x^3-1)}+\frac{\left(1-\frac13\right)}{(-1)1!}\left(-\frac1{3(-1)}\right)\left\{\frac{(-1)}2\ln(x^2-(-1)x+1)\right.\\ &\quad\left.+\sqrt3\tan^{-1}\left(\frac{\sqrt3}{2x-(-1)}\right)-(-1)\ln(x+(-1))\right\}+C\\ &=-\frac{x}{3(x^3-1)}-\frac29\left\{-\frac12\ln(x^2+x+1)\right.\\ &\quad\left.+\sqrt3\tan^{-1}\left(\frac{\sqrt3}{2x+1}\right)+\ln(x-1)\right\}+C\end{align}$$ Also checked. Lets see... my formulas have problems with discontinuities and domains so I should probably change $$\tan^{-1}\left(\frac{\sqrt3}{2x\pm1}\right)$$ to $$\frac{\pi}2-\tan^{-1}\left(\frac{2x\pm1}{\sqrt3}\right)$$ And also $\ln(x\pm1)$ t0 $\ln\lvert x\pm1\rvert$ to fix these problems.

Perhaps a little long-winded but I wanted to sum the recurrence relation in case a higher power of the denominator was assigned next week.

Answer 4

Hint:

Simplify

$$\int{\frac{1}{(x^3+1)^2}}dx=\frac x{3(x^3+1)}+\frac13\int\frac2{x^3+1}dx$$

Then, decompose

$$\frac2{x^3+1}=\frac1{x+1}+\frac{1}{x^2-x+1}- \frac{x^2}{x^3+1}$$

Answer 5

$$ \begin{aligned} I &=\int \frac{1}{\left(x^{3}+1\right)^{2}} d x \\ &=-\frac{1}{3} \int \frac{1}{x^{2}} d\left(\frac{1}{x^{3}+1}\right) \\ &=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int \frac{1}{x^{3}\left(x^{3}+1\right)} d x \\ &=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int\left(\frac{1}{x^{3}}-\frac{1}{x^{3}+1}\right) d x \\ &=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}+\frac{1}{3 x^{2}}+\frac{1}{3} \int \frac{d x}{x^{3}+1} \end{aligned} $$

By my post,

$$ \therefore I=\frac{1}{18}\left[\frac{6 x}{\left(x^{3}+1\right)}+\ln \left|\frac{(1+x)^{2}}{1-x+x^{2}}\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]+C $$