Integration by Parts. Stuck on substitution step Trying to solve $$\int 27x^3(9x^2+1)^{12} dx$$
I know the process and formula of integration by parts. When I set $u = 9x^2 + 1$, $du = 18x dx$. I am stuck on the next step as 18x does not line up with the $27x^3$. 
 A: I think you don't need to use integration by parts. Substitution is enough. Let $u=9x^2+1$. Then $du=18xdx$ or $xdx=\frac1{18}du$, $x^2=\frac19(u-1)$ and hence
\begin{eqnarray}
\int 27x^3(9x^2+1)^{12} dx&=&27\int x^2(9x^2+1)^{12} xdx\\
&=&27\cdot\frac19\cdot\frac{1}{18}\int(u-1)u^{12}du\\
&=&\frac16\left(\frac1{14}u^{14}-\frac1{13}u^{13}\right)+C\\
&=&\frac16\left(\frac1{14}(9x^2+1)^{14}-\frac1{13}(9x^2+1)^{13}\right)+C.
\end{eqnarray}
A: As you saw, substitution is not the way to go, because the powers do not work out nicely. Try reducing your problem to an easier one to use substitution on by taking
$$
u=27x^2\\
\mathrm dv=x(9x^2+1)^{12}
$$
in the formula 
$$
\int u\mathrm dv=uv-\int v\mathrm du
$$
A: With integration by parts, formula is
$$\int u dv = uv - \int v du$$
When you set $u = 9x^2+1$, in order to apply integration by parts, you must let $dv = \left(27x^{3}(9x^2+1)^12\right)/(9x^2+1)$. In this case, it might be easier to let $u$ be something else.
Regardless, once you've chosen your values of $u$ and $dv$, you can use them (and the corresponding $du$ and $v$ you get by taking derivatives and antiderivatives, respectively), to plug them into the right side of the above inequality.
What you are hoping for is to choose $u$ and $dv$ such that $\int vdu$ is an easier integral for you to solve.
