Integration problem: $\int x^{2} -x 4^{-x^{2}} dx$ I need to integrate $$\int x^{2} -x 4^{-x^{2}} dx$$ 
and I know the answer I got is wrong. However, I can't figure out where I went wrong. These steps I took:

Appreciate any help!
 A: $$\begin{align}
\int x^{2} - x4^{-x^{2}} dx &= \int x^{2} dx - \int x4^{-x^{2}} dx \\
&= \frac{x^{3}}{3} + \int -x4^{-x^{2}} dx \\
\end{align}$$
notice that 
$$\frac{d}{dx} 4^{-x^{2}} = -2x\cdot \ln4 \cdot 4^{-x^{2}}$$
So if 
$$\begin{align}
u &= 4^{-x^{2}} \\
\implies du &= -2x\cdot \ln4 \cdot 4^{-x^{2}} dx \\
\implies \frac{du}{2 \ln4} &= -x 4^{-x^{2}} dx
\end{align}$$
So our integral becomes
$$\begin{align}
\frac{x^{3}}{3} + \int -x4^{-x^{2}} dx &= \frac{x^{3}}{3} + \int \frac{du}{2 \ln4} \\
&= \frac{x^{3}}{3} + \frac{u}{2 \ln4} + C \\
&= \frac{x^{3}}{3} + \frac{4^{-x^{2}}}{2 \ln4} + C \\
\end{align}$$
A: I would approach it as follows:
\begin{align}
\int \left(x^2-\frac{x}{4^{x^2}}\right)\,dx &= \int x^2\,dx - \int 4^{-x^2}x \,dx\\[1em]
               &= \frac{x^3}{3}-\int 4^{-x^2}x\,dx\\[1em]
               &= \frac{x^3}{3}+\frac{1}{2}\int 4^u\,du\tag{$u=-x^2; du=-2x\,dx$}\\[1em]
               &= \frac{2^{2u-1}}{\log(4)}+\frac{x^3}{3}+C\\[1em]
               &= \frac{x^3}{3}+\frac{2^{-2x^2-1}}{\log(4)}+C\tag{subst. back $u=-x^2$}\\[1em]
               &= \frac{x^3\log(4)+3\cdot 2^{-2x^2-1}}{\log(64)}+C,
\end{align}
As you can see, the real trick is to use $u$-substitution effectively, which I imagine is the point of this exercise.
