# Recreational integral problem: $\int_0^1 (x+\sqrt{x^3-1})^{2018}dx$

I have a (recreational) integral problem any interested inhabitants of MSE to enjoy:

Evaluate the following integral: $$\int_0^1 \bigg(x+\sqrt{x^3-1}\bigg)^{2018}dx$$

I promise, it's not from an ongoing math competition (I just put the $2018$ in there to be funny). If you don't believe me, you can try solving the problem with $1,000,000$ (or any large even number, really) instead of $2018$.

Cheers!

• @tired What do you mean? It's defined the way the cube root has always been defined on real numbers... May 7, 2018 at 23:40
• @tired How so? Try plugging in $1/2$... May 7, 2018 at 23:45
• @tired I don't think so. For every real number $a$, there exists exactly one real number $b$ such that $b^3=a$. We define the cube root of $a$ to be equal to $b$. This should always work out just fine, since the function $f(x)=x^3$ is a bijective function from reals to reals. May 7, 2018 at 23:45
• @tired wolframalpha differentiate between $x^{1/3}$ and $\sqrt{x}$. The first one is understood as a complex valued function, the second no. Take a look here May 7, 2018 at 23:47
• @tired I guess you're too tired to continue. May 8, 2018 at 0:42

Write $$I = \int_0^1 \bigg(x+\sqrt{x^3-1}\bigg)^{2018}dx.$$ Making the substitution $u^3 + x^3 = 1$, we get $$I = \int_0^1 \frac{u^2}{\sqrt{1-u^3}^2}\bigg(u+\sqrt{u^3-1}\bigg)^{2018}\,du.$$ (The fact that $2018$ is even is important here, as it removes a minus sign inside the parentheses). Hence $$2I = \int_0^1 \left(\frac{u^2}{\sqrt{1-u^3}^2}+1\right)\bigg(u+\sqrt{u^3-1}\bigg)^{2018}\,du.$$ But, if $v = u + \sqrt{u^3-1}$ then $\frac{dv}{du} = \frac{u^2}{\sqrt{1-u^3}^2}+1$, which is exactly the left factor. So, $$2I = \int_{-1}^1 v^{2018}\,dv=\frac{2}{2019}$$ Hence $I = \frac{1}{2019}$.
• how you get your $2I$? I have that $$\frac{u^2}{\sqrt{1-u^3}^2}+1=\frac{u^2+\sqrt{1-u^3}^2}{\sqrt{1-u^3}^2}\neq \frac{2u^2}{\sqrt{1-u^3}^2}$$ May 8, 2018 at 1:57
• Wow, elegant!$\phantom{}$ May 8, 2018 at 4:03
• @B.Mehta Nicely done; this is just how I did it! In general, I've discovered this fancy integral trick: if $E$ is an even function, $\theta$ is an involution, and $a$ is a constant, then $$\int_a^{\theta(a)} E(x-\theta(x))dx=\int_0^{\theta(a)-a}E(x)dx$$ May 8, 2018 at 22:59