I have a (recreational) integral problem any interested inhabitants of MSE to enjoy:
Evaluate the following integral: $$\int_0^1 \bigg(x+\sqrt[3]{x^3-1}\bigg)^{2018}dx$$
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!
Write $$I = \int_0^1 \bigg(x+\sqrt[3]{x^3-1}\bigg)^{2018}dx.$$ Making the substitution $u^3 + x^3 = 1$, we get $$I = \int_0^1 \frac{u^2}{\sqrt[3]{1-u^3}^2}\bigg(u+\sqrt[3]{u^3-1}\bigg)^{2018}\,du.$$ (The fact that $2018$ is even is important here, as it removes a minus sign inside the parentheses). Hence $$2I = \int_0^1 \left(\frac{u^2}{\sqrt[3]{1-u^3}^2}+1\right)\bigg(u+\sqrt[3]{u^3-1}\bigg)^{2018}\,du.$$ But, if $v = u + \sqrt[3]{u^3-1}$ then $\frac{dv}{du} = \frac{u^2}{\sqrt[3]{1-u^3}^2}+1$, which is exactly the left factor. So, $$2I = \int_{-1}^1 v^{2018}\,dv=\frac{2}{2019}$$ Hence $I = \frac{1}{2019}$.