Where am I going wrong integrating $\int x^2(x^3-6)^{34} dx$ by parts? I was helping a friend with homework and was given the fairly straightforward integral:
$$\int x^2(x^3-6)^{34} dx$$
The answer the homework checker accepts as correct baffles me since it has one term with one independent degree:
$$\frac1{105}(x^3-6)^{35}+ C$$
That in no way whatsoever resembles my answer:
$$(x^3-6)^{34}(1/3x^3) - \frac {17}{21} x^{42} + \frac {102}{19} x^{38} + C$$
I feel like there is some weird trick that allows us to avoid integrating by parts, but they wanted us to use integration by parts with $u=x^3-6$. These are the steps I used:
$$
\begin{align}
&u = x^3 - 6\\
&w = u^{34}\\
&\frac {d}{dx} w = \frac {dw}{du} \frac {du}{dx} dx\\
&\cdots = (34u^{33})(3x^2)dx = 102x^{38} - 612x^{35}\\
&dv = x^2 dx\\
&v = 1/3 x^3 + C\\
&I = wv - \int v dw\\
&\cdots= (x^3-6)^{34}(1/3x^3) - I_2\\
&I_2 = \int (1/3 x^3)(102x^{38} - 612x^{35}) dx
\end{align}$$
Regardless of anything else, I can't figure out how you are supposed to have $1$ term with one degree, $35$, in our answer.

EDIT: Just adding what made the accepted answer make sense to me, $$x^2 u^{34} dx = C(u^{34} du)$$ where $C$ is a constant because I wanted to emphasize the general form.
 A: Integration by parts is hardly needed.
We can use substitution: Let $u = (x^3 - 6)$.  So $du = 3x^2\,dx$. Given $$\int x^2(x^3-6)^{34} dx = \frac 13 \int 3x^2(x^3-6)^{34} dx,$$
under substitution, the integral becomes
$$\frac 13\int u^{34} du.$$
Can you take it from here?
A: 
$$\int x^2(x^3-6)^{34} dx$$


but they wanted us to use integration by parts with u=x^3-6.

I certainly agree with amwhy's answer.  Integration by parts is an unnecessary complication, since the $u = \left(x^3 - 6\right)$ substitution implies (informally) that $du = 3x^2 dx.$
However, this begs the question: what was the problem composer's intended solution, under the assumption that you (somehow) combine $u = (x^3 - 6)$ with integration by parts?
So, I will take a crack at what the problem composer might have intended.
For convenience, I will omit displaying any constants of integration.
I will use the idea that
$$\left(\int a ~db\right) = ab - \left(\int b ~da\right). \tag1 $$
Set $\displaystyle I = \int x^2(x^3-6)^{34} dx.$
Set $u = (x^3 - 6).$
Set $a = u^{34}.$ 
Set $db = x^2~dx.$
Then, $\displaystyle I = \int a ~db.$
Following the formula in (1) above,
$\displaystyle I = ab - \int b ~da.$
Since $a = u^{34},$ you have that $da = 34u^{33} du.$
Since $db = x^2 ~dx$, you have that
$$b = \frac{x^3}{3} = \frac{1}{3} (u + 6). \tag2 $$

Edit
I think that arguments could be made, either way that:

*

*This whole approach is daffy, and a substitution attempt should be your first try.  Then, integration by parts would not be used for this problem.


*In rebuttal, in addition to reinforcing the correct use of integration by parts, you have that the new Math student is accidentally discovering the convenient result in (2) above, by the pedestrian method of integration by parts.  So, there is the suggestion that if an integration problem does not seem to immediately bend to a substitution, that integration by parts might reveal an insight.

So, $\displaystyle I = u^{34}\left[\frac{1}{3} (u + 6)\right] - J
 ~=~ \left[\frac{u^{35}}{3} + 2u^{34}\right] - J$,
where $\displaystyle J = \int 34u^{33}\left[\frac{1}{3} (u + 6)\right] ~du = \int \left[\frac{34u^{34}}{3} + 68u^{33}\right] ~du.$
You have that $~\displaystyle 
J = \frac{34u^{35}}{105} + 2u^{34}.$
Therefore,
$~\displaystyle I = \left[\frac{u^{35}}{3} + 2u^{34}\right] - 
\left[\frac{34u^{35}}{105} + 2u^{34}\right] = \frac{u^{35}}{105}
 ~=~ \frac{(x - 6)^{(35)}}{105}.$
