$\int_0^t (t-r)^{1/3} r^{5/3} dr$ The integral $$\int_0^t (t-r)^{1/3} r^{5/3} dr$$ came up in Miller's Introduction to Differential Equations (1987), (Sec 6.6, Convolutions, in the chapter on Laplace transformations, problem 5a on p. 328). Given the section, convoluted functions multiply under the Laplace transform, and etc. Okay, fine. But taken out of context as just another integral, I was stumped. If there were squared terms in the parentheses I could try trig substitutions. If the exponents were integers, fine. If it was just a single term with a fractional exponent, and not multiplied by that difference, okay. But this general form with fractional exponents, I'm not sure what to do. How do I solve this and other integrals like it?
I'm going to try sticking a second question in here. Miller integrates his convolutions from 0 to t, other people seem to go from minus infinity to infinity. What's with that?
(It's been a long time since I've used LaTeX, and I'm new to MathJax on the web, so I'm not sure how this is going to turn out.)
 A: Make the substitution $r=ts$ (I'm assuming $t>0$), so that the integral becomes
\begin{align}
\int_0^t(r-t)^{1/3}r^{5/3}\,dr&=t^3\int_0^1s^{5/3}(1-s)^{1/3}\,ds\\
&=t^3B\left(\frac{5}{3}+1,\frac{1}{3}+1\right)\\
&=t^3B\left(\frac{8}{3},\frac{4}{3}\right),
\end{align}
where $B(x,y):=\int_0^1s^{x-1}(1-s)^{y-1}\,ds$ is the Beta function (defined for $x,y\in\Bbb{C}$ such that $\text{Re}(x),\text{Re}(y)>0$). Now, it is a "standard" fact that the Beta function is related to the Gamma function via the identity $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. So, in our case, we have
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{\Gamma\left(\frac{8}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{8}{3}+\frac{4}{3}\right)}=
\frac{\Gamma\left(\frac{8}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(4\right)}
\end{align}
Now, a fundamental property of the Gamma function is that $\Gamma(1+z)=z\Gamma(z)$ (a factorial-like property). Therefore, the above reduces to
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{\left[\frac{5}{3}\cdot\frac{2}{3}\cdot \Gamma\left(\frac{2}{3}\right)\right]\cdot
\left[\frac{1}{3}\cdot \Gamma\left(\frac{1}{3}\right)\right]}{3\cdot 2\cdot \Gamma(1)}=
\frac{5}{81}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma(1)}
\end{align}
Now, another fact about the Gamma function is that $\Gamma(1)=1$ (trivial from the definitions), and the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$. Using this with $z=\frac{1}{3}$, we get
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{5}{81}\cdot\frac{\frac{\pi}{\sin\left(\frac{\pi}{3}\right)}}{1}=\frac{10\pi}{81\sqrt{3}}.
\end{align}
Therefore, our original integral is $t^3\cdot \frac{10\pi}{81\sqrt{3}}$.

Now, in general, integrals with fractional/irrational exponents are not easy at all.  We have special functions like the Beta and Gamma and Zeta functions to deal with them, and we have exploit nice identities involving them, and one of the many ways of proving these identities uses several ideas from complex analysis (eg infinite products, contour integration etc). As always with integration, there is no cookie-cutter recipe; you just have to deal with each case/class of integrals separately.

Edit: Alternative Approach.
Rewrite the integral as
\begin{align}
\int_0^1s^{5/3}(1-s)^{1/3}\,ds&=\int_0^1\left(\frac{s}{1-s}\right)^{5/3}(1-s)^2\,ds
=\int_0^{\infty}\frac{x^{5/3}}{(1+x)^4}\,dx\\
&=\int_0^{\infty}x^{2/3}\frac{x}{(1+x)^4}\,dx
\end{align}
where we made the substitution $x=\frac{s}{1-s}$. Now, this last integral is again of a "standard form" found in complex analysis texts, which means you can integrate along a pacman contour/keyhole contour, as I explain in detail in this post (with $\alpha=\frac{2}{3}, R(x)=\frac{x}{(1+x)^4}$). Now, you can have fun calculating residues (this is essentially what I meant in my previous comment regarding proving identities using complex analysis and contour integrals).
A: Substutute  $r=\frac {tx^3}{1+x^3}$
\begin{align}
I=\int_0^t(r-t)^{1/3}r^{5/3}\,dr&=3t^3\int_0^\infty 
\frac{x^7}{(1+x^3)^4}dx
\end{align}
and then apply the reduction $\int_0^\infty 
\frac{x^m}{(1+x^3)^n}dx= J_{m,n}= \frac{m-2}{3(n-1)} J_{m-3,n-1}$
\begin{align}
I= &\frac{5t^3}9\int_0^\infty 
\frac{x}{(1+x^3)^2}dx
= \frac{5t^3}9 \int_0^\infty \frac1{3x}d\left(\frac{x^3}{1+x^3} \right)\\
=&\frac{5t^3}{27} \int_0^\infty \frac{x}{1+x^3}\overset{x\to\frac1x}{dx}=\frac{5t^3}{54} \int_0^\infty \frac{1+x}{1+x^3}{dx}\\
=&\frac{5t^3}{54} \int_0^\infty \frac{1}{1-x+x^2}{dx}
=\frac{10\pi t^3}{81\sqrt3}
\end{align}
