trouble with double/iterated integration of $\int^1_0[\int^1_0v(u+v^2)^4du]dv$ I have:
For $\int^1_0v(u+v^2)^4du$:
u substitution (using x instead since there's a u in there already) with $x=(u+v^2), dx/du=1$
$v\int^1_0x^4=v\frac{1}{5}x^5=v\frac{(u+v^2)^5}{5}-\frac{v(0+v^2)^5}{5}|^1_0=\frac{(v+v^3)^5}{5}-\frac{v^{15}}{5}$
then
$ \frac{1}{5}[\int^1_0(v+v^3)^5dv-\int^1_0v^{15}dv]$
used substitution again with $x= (v+v^3),dx/du=(1+3v^2)$ for the first one
$\int^1_0\frac{x^5dx}{(1+3v^2)}->\frac{1}{1+3v^2}\int^1_0x^5dx=\frac{1}{1+3v^2}\frac{1}{6}x^6|^1_0->\frac{1}{1+3v^2}\frac{1}{6}(v+v^3)^6|^1_0$
$= \frac{1}{24}*1-0$
for the second:
$\int^1_0 v^{15}=\frac{1}{16}v^{16}|^1_0=\frac{1}{16}$
$\frac{1}{5}[\frac{1}{24}-\frac{1}{16}] = -\frac{1}{240}$
The answer is supposed to be 31/30.

An alternate approach would be to change the order of integration. This will greatly reduce the difficulty of the integral. As the limits of the inner integral do not depend upon $v$ this trivially becomes:
$$\int^1_0[\int^1_0v(u+v^2)^4\ dv]\ du$$
$$=\int^1_0\left[\frac{1}{10}\bigg((u+v^2)^5\bigg)_0^1\right]\ du$$
$$=\int^1_0\frac{1}{10}\left[(u+1)^5-(u+0)^5\right]\ du$$
$$=\frac{1}{10}\int^1_0(u+1)^5-u^{5}\ du$$
$$=\frac{1}{10}\bigg(\frac{1}{6}(u+1)^6-\frac{1}{6}u^6\bigg)_0^1$$
$$=\frac{1}{10}\bigg(\frac{1}{6}(1+1)^6-\frac{1}{6}1^6-\left(\frac{1}{6}(1+0)^6-\frac{1}{6}0^6\right)\bigg)_0^1$$
$$=\frac{31}{30}$$
