indefinite integral by substitution I know we are supposed to post as far as we get on a problem in order not to waste people's time, and I understand the reason for that. But I really couldn't get very far on this problem. I am not just looking for an answer, I need to understand how to solve this for my final. Thanks.
My problem is $$\int((6z)^5+4(6z)^2)((6z)^3+1))^{12}dz$$.
I tried to solve by using $u$ substitution, to no avail. I set $6z$ as $u$, and got $$\frac {du}{dz}=6$$
I then moved the $dz$ to the other side, and multiplied both sides by $2/3$ in order to get 
$$\frac {2\,du}{3}=4\,dz$$ I did this in order to try to match with the $4$ in front of the $$(6z)^2$$
I then attempted to substitute the $6z$ with $u$, but that didnt help me at all.
I'm really lost on this one, can someone please point me in the right direction?
Thanks 
 A: Just set $u=6z$ as you did, so $du=6\,dz$ and $dz=\dfrac{du}6$. Then your integral is
$$\int (u^5+4u^2)(u^3+1)^{12}\cdot \frac 16\,du$$
Can you continue from there?
A: Starting from Rory Daulton's answer (forgetting the factor $6$), you could (being brave !) expand $(u^3+1)^{12}$ and multiply the result by $(u^5+4u^2)$ and expand again. This is workable and would lead, for the integrand, to the nice monster $$u^{41}+16 u^{38}+114 u^{35}+484 u^{32}+1375 u^{29}+2772 u^{26}+4092 u^{23}+4488
   u^{20}+3663 u^{17}+2200 u^{14}+946 u^{11}+276 u^8+49 u^5+4 u^2$$
However, you can make this nicer and more compact writing the integrand as $$u^5(1+u^3)^{12}+4u^2(1+u^3)^{12}$$ Now, remember the binomial theorem $$(1+x)^n=\sum_{k=0}^n \binom{n}{k} x^k$$ This makes $$(1+u^3)^{12}=\sum_{k=0}^{12} \binom{12}{k} u^{3k}$$ So, the integrand write now $$\sum_{k=0}^{12} \binom{12}{k} u^{3k+5}+4\sum_{k=0}^{12} \binom{12}{k} u^{3k+2}$$ and so the integral is $$\sum_{k=0}^{12} \frac{\binom{12}{k} }{3 k+6}u^{3 k+6}+4\sum_{k=0}^{12} \frac{\binom{12}{k} }{3 k+3}u^{3 k+3}$$
