# Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ [closed]

I've been starting to work on problems with difficult integrals, and I came to one of my problems where I end up having to integrate $$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$ Wolfram seems to time out on integrals like these, so I decided to tackle this integral by myself.

It seems like to me that there should be a meaningful integration by parts that occurs here, in particular a relationship that could be developed between $x^5$ and $\frac{(1-x)^4}{1+x^3}.$ The problem is, I'm having some difficulties with the difficulty of building a meaningful integral of $dv = \frac{(1-x)^4}{ 1+x^3} dx.$ any recommendations on how to go about starting with direction on this problem?

## closed as off-topic by user99914, user91500, Claude Leibovici, Watson, tiredSep 13 '16 at 12:03

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• What makes you say this integral is very interesting? – Gerry Myerson Sep 13 '16 at 7:24
• for me this integral looks disgusting – tired Sep 13 '16 at 12:03

Using the binomial theorem, expand $(1-x)^4$ in the numerator: $$I = \int_0^1 \frac{x^5(1-x)^4}{1+x^3}\,dx = \int_0^1 \frac{x^5(1 - 4x + 6x^2 - 4x^3 + x^4)}{1+x^3}\,dx \\ = \int_0^1 \frac{x^5 - 4x^6 + 6x^7 - 4x^8 + x^9}{1+x^3}\,dx.$$ Using polynomial long division, $$I = \int_0^1 \left(x^6 - 4x^5 + 6x^4 - 5x^3 + 5x^2 - 6x + 5 + \frac{-5x^2 + 6x - 5}{x^3 + 1}\right)\,dx.$$ At this point let's focus on the term $\frac{-5x^2 + 6x - 5}{x^3 + 1}$. Notice that $x^3 + 1 = (x+1)(x^2-x+1),$ and perform partial fraction decomposition: $$\int_0^1\frac{-5x^2 + 6x - 5}{x^3 + 1}\,dx = \int_0^1\left(\frac{x+1}{3(x^2-x+1)} - \frac{16}{3(x+1)}\right)\,dx.$$ Put $\alpha = 3/2$, $\beta^2 = \alpha^2/3$, $u = x-1/2$, and let's focus on the term $\frac{x+1}{3(x^2-x+1)}$. Completing the square in the denominator: $$\int_0^1\frac{(x-\tfrac{1}{2}) + \tfrac{3}{2}}{3(x-\tfrac{1}{2})^ 2 + \tfrac{9}{4}}\,dx = \int_{-1/2}^{1/2}\frac{u + \alpha}{3u^2 + \alpha^2}\,du = \frac{1}{3}\int_{-1/2}^{1/2}\frac{\alpha}{u^2 + \tfrac{\alpha^2}{3}}\,du \\= \frac{2\alpha}{3}\int_{0}^{1/2}\frac{1}{u^2 + \beta^2}\,du = \frac{1}{\beta}\,\arctan{\frac{u}{\beta}}\,\Big|_{0}^{1/2} = \frac{2}{\sqrt{3}}\,\arctan{\frac{1}{\sqrt{3}}} = \frac{2}{\sqrt{3}}\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}},$$ where we made use of the facts that $2\alpha/3 = 1$, $f(u) = u/(3u^2 + \alpha^2)$ is odd, $g(u) = 1/(3u^2+\alpha^2)$ is even, and our region of integration is symmetric about $u = 0$. The integral $\int_0^1 \frac{16}{3(x+1)}\,dx$ should be straightforward, and of course the integral of the polynomial portion is easy too.
In general, any integral of the form $\int_{0}^{1}\frac{p(x)}{1+x^3}\,dx$ with $p(x)\in\mathbb{Q}[x]$ can be written as a linear combination with rational coefficients of $1$ and $$I_1=\int_{0}^{1}\frac{x}{1+x^3}\,dx=\frac{1}{9} \left(\pi \sqrt{3} -3\log 2\right),\qquad I_2= \int_{0}^{1}\frac{x^2}{1+x^3}\,dx=\frac{1}{3}\log(2)$$ $$I_0=\int_{0}^{1}\frac{1}{1+x^3}\,dx = \frac{1}{9} \left(\pi\sqrt{3} +3\log 2\right)$$ and we just need to perform a polynomial division betwen $p(x)$ and $(1+x^3)$ to compute the original integral. Notice that $I_2$ is trivial and $$I_1+I_0 = \int_{0}^{1}\frac{dx}{1-x+x^2}=\frac{2\pi}{3\sqrt{3}},$$ $$I_1-I_0 = \int_{0}^{1}\frac{x-1}{x^3+1}=-\sum_{n\geq 0}\left(\frac{1}{6n+1}-\frac{1}{6n+2}-\frac{1}{6n+4}+\frac{1}{6n+5}\right)=-\frac{2\log 2}{3}$$ are quite simple to prove.
For starters, you will want to divide the polynomials, since the degree on top is greater than the degree on bottom. Then you can take advantage of the fact that the bottom $1+x^3=(1+x)(1-x+x^2)$ and perform partial fractions accordingly.