Does this one require integration by parts? $$\int_{0}^{1} x \sqrt{1+8x^2}\,dx$$
I tried 
$u=x$
$\text{d}u=\text{d}x$
$\text{d}v = (1+8x^2)^\frac{1}{2}$
but I'm not sure how to get $v$ by integrating $dv$, since $u$-sub doesn't work.
Barking up the wrong tree?
 A: (Joke) We will use integration by parts!
Let $u=\sqrt{1+8x^2}$ and $dv=x\,dx$. Then $du=\frac{8x}{\sqrt{1+8x^2}}\,dx$ and we can let $v=\frac{x^2}{2}+\frac{1}{16}=\frac{1+8x^2}{16}$. Thus our integral $I$ is given by
$$I=\left.\frac{1}{16}(1+8x^2)^{3/2}\large\right|_0^1-\int_0^1 \frac{1}{2}x\sqrt{1+8x^2}\,dx=\left.\frac{1}{16}(1+8x^2)^{3/2}\large\right|_0^1-\frac{I}{2}.$$
It follows that
$$\frac{3}{2}I=\frac{1}{16}(27-1),$$
so $I=\frac{13}{12}$. 
Remark: After learning about integration by parts, some students develop a great fondness for it, and may miss substitutions that would have been routine a few days earlier. 
A: With the not so-trivial-substitution $x=\frac{\sinh(z)}{2\sqrt{2}}$ the integral becomes:
$$ \frac{1}{8}\int_{0}^{\text{arcsinh}(2\sqrt{2})}\sinh(u)\cosh(u)^2\,du=\left.\frac{\cosh^3(u)}{24}\right|_{0}^{\text{arccosh}(3)}=\frac{27-1}{24}=\color{red}{\frac{13}{12}}.$$
A: Hint:
Substitute:
$$
1+8x^2=u \quad \Rightarrow \quad 16xdx=du \quad \Rightarrow \quad x dx=\frac{1}{16}du
$$
