# How to evaluate the integral: $\int_{0}^{1}\frac{dx}{1+x^{3}}$

I am a bit lost on how to evaluate the integral: $$\int_{0}^{1}\frac{dx}{1+x^{3}}$$

I tried the substitution: $y=x^{3}$, but I got a more complicated integrand. Any ideas?

• Try partial fraction decomposition first. – Qiaochu Yuan Apr 12 '12 at 15:43

There's a factorization of a sum of two cubes: $$1+x^3 = (1+x)(1-x+x^2)$$ This quadratic polynomial is irreducible unless you allow imaginary numbers since if you write it as $ax^2+bx+c$, it turns out that $b^2-4ac<0$. So $$\frac{1}{1+x^3} = \frac{1}{(1+x)(1-x+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1-x+x^2}$$ As usual with partial fractions, you then find $A$, $B$, and $C$. The integral of $A/(1+x)$ is easy. The other one you can start with a substitution: $u=x^2-x+1$, $du = (2x-1)\,dx$. So $$(Bx+C)\,dx = \frac B 2 \left(2x + \frac{-2}{B} \right)\,dx + (\text{some constant (find it!)})\,dx$$ Then you need $$\frac B 2 \int \frac{2x-1}{x^2-x+1}\,dx$$ and you can do that by using the substitution.
Finally, you need $$\int \frac{\text{constant}}{x^2-x+1}\,dx.$$ Complete the square: $$x^2 - x + 1 = \left(x^2 - x + \frac 1 4\right) + \frac 3 4 = \left(x - \frac 1 2\right)^2 + \frac 3 4$$ We'd like $1$ where we see $3/4$, so that it will look like the derivative of the arctangent. So $$\left(x - \frac 1 2\right)^2 + \frac 3 4 = \frac 3 4 \left( \frac 4 3\left(x - \frac 1 2\right)^2 + 1 \right) = \frac 4 3 \left(\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1\right) = \frac 4 3 (w^2 + 1)$$ and then $dw = \dfrac{2}{\sqrt{3}}\,dx$
Finally you have $$(\text{constant})\cdot \int \frac{dw}{w^2 + 1} = \text{constant}\cdot\arctan(w) + c$$ and then you convert it back to a function of $x$.