Partial Fractions = A Partial Answer

First of all,

$1+x^3 = (1+x)(x^2-x+1)$

Hence, we have to evaluate,

$\displaystyle\int\limits_0^1\frac{1}{(1+x)(x^2-x+1)} dx$

Using Partial Fractions, we can split the integral term as,

$\displaystyle\frac{1}{(1+x)(x^2-x+1)} = \frac{A}{1+x}+\frac{Bx+C}{x^2-x+1}$

$\displaystyle\Rightarrow (A+B)x^2+(B+C-A)x+(C+A)=1$

Comparing coefficients and solving the equations, we get,

$\displaystyle A=\frac{1}{3}, B=\frac{-1}{3}, C=\frac{2}{3}$

Substituting back in the equation,

$\displaystyle \frac{1}{3}\int\limits_0^1\frac{dx}{1+x} + \frac{1}{3}\int\limits_0^1\frac{-x+2}{x^2-x+1}dx$

$\displaystyle\frac{1}{3} ln2 -\frac{1}{6}\int\limits_0^1\frac{2x-4}{x^2-x+1}dx$

$\displaystyle\frac{1}{3} ln2 -\frac{1}{6}\int\limits_0^1\frac{2x-1}{x^2-x+1}dx + \frac{1}{2}\int\limits_0^1\frac{dx}{x^2-x+1}$

$\displaystyle\frac{1}{3} ln2 - 0 + \frac{2}{\sqrt{3}}(\frac{\pi}{3})$

$\displaystyle\frac{1}{3} ln2 + \frac{2}{\sqrt{3}}(\frac{\pi}{3})$

Here, I dont know if I went wrong somewhere or the question went wrong, but either way, to get the answer, we have to replace the " $2$ " in the second term with a " $1$ "

$\displaystyle\frac{1}{3} ln2 + \frac{1}{\sqrt{3}}(\frac{\pi}{3})$

$\displaystyle\frac{1}{\sqrt{9}} ln2 + \frac{1}{\sqrt{27}} \pi$

Hence,

$a=27$ and $b=9$

$\Rightarrow a+b = \boxed{36}$

To simplify notation, I will omit the limits of integration until the end, first focusing on finding the indefinite integral. (I will also omit the constant of integration throughout.) We will begin by separating the expression into two, using the method of partial fractions and the fact that $x^3+1=(x+1)(x^2-x+1)$ :

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

The first integral will not be difficult to evaluate, but the second one is tougher. If we try the natural logarithm route again, we find that we need a constant multiple of $2x-1$ in the numerator, which we don't have. But we can do a second separation to get one:

$\frac{1}{3}\int\frac{x-2}{x^2-x+1}dx = \frac{1}{6}\int\left(\frac{2x-1}{x^2-x+1}-\frac{3}{x^2-x+1}\right)dx \\ = \frac{1}{6}\int\frac{2x-1}{x^2-x+1}dx-\frac{1}{6}\int\frac{3}{x^2-x+1}dx$

So now we've split the original integral into three:

$\frac{1}{3}\int\frac{1}{x+1}dx - \frac{1}{6}\int\frac{2x-1}{x^2-x+1}dx + \frac{1}{6}\int\frac{3}{x^2-x+1}dx$

As mentioned, the first two are straightforward to integrate:

$\frac{1}{3}\ln(x+1)-\frac{1}{6}\ln(x^2-x+1)+\frac{1}{6}\int\frac{3}{x^2-x+1}dx$

In the third integral, we can complete the square in the denominator and do a u -substitution:

$\frac{1}{6}\int\frac{3}{x^2-x+1}dx = \frac{1}{6}\int\frac{3}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx \\ = \frac{1}{6}\int\frac{1}{\frac{1}{3}\left(x-\frac{1}{2}\right)^2+\frac{1}{4}}dx = \frac{1}{6}\int\frac{4}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}dx$

Let $u=\frac{2x-1}{\sqrt{3}}$ . Then $du=\frac{2}{\sqrt{3}}dx$ , and this integral is

$\frac{1}{6}\int\frac{4}{u^2+1}\cdot\frac{\sqrt{3}}{2}du = \frac{\sqrt{3}}{3}\int\frac{1}{u^2+1}du = \frac{\sqrt{3}}{3}\tan^{-1}u = \frac{\sqrt{3}}{3}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right).$

So our complete antiderivative is

$\frac{1}{3}\ln(x+1)-\frac{1}{6}\ln(x^2-x+1)+\frac{\sqrt{3}}{3}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right).$

Now bringing back the limits of integration and evaluating, we have

$\left(\frac{1}{3}\ln 2 - \frac{1}{6}\ln 1+\frac{\sqrt{3}}{3}\tan^{-1}\frac{1}{\sqrt{3}} \right)-\left(\frac{1}{3}\ln 1 - \frac{1}{6}\ln 1+\frac{\sqrt{3}}{3}\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \right),$

which simplifies to $\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9}$ . Putting this in the desired form gives $\frac{\pi}{\sqrt{27}}+\frac{1}{\sqrt{9}}\ln 2$ , so the answer is $27+9=\boxed{36}$ .