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.

Note that $v(1+v^2)^5\ne (v+v^3)^5$.  Rather, we have upon integrating with respect to $u$,
$$ \int_0^1 v(u+v^2)^4\,du=\frac v5 \left( (1+v^2)^5-v^{10} \right)$$
Now, note that the integration over $v$ is facilitated by the substitution $w=1+v^2$.  This is left as an exercise for the reader.
