Evaluate $\int_{0}^{1}4x^3\left(\frac{d^2}{dx^2}(1-x^2)x^5\right)dx$ $$\int_{0}^{1}4x^3\left(\dfrac{d^2}{dx^2}(1-x^2)x^5\right)dx$$
$$\int_{0}^{1}4x^3\left(\dfrac{d}{dx}(5x^4-7x^6)\right)dx$$
$$\int_{0}^{1}4x^3\left(20x^3-42x^5\right)dx$$
$$8\int_{0}^{1}10x^6-21x^8 dx$$
$$8\left(\dfrac{10x^7}{7}-\dfrac{21x^9}{9}\right)_{x=1}$$
$$8\left(\dfrac{10}{7}-\dfrac{21}{9}\right)=8\cdot\dfrac{90-147}{63}=-\dfrac{152}{21}$$
But actual answer is $2$. What am I missing here?

Although you've gotten to the point of knowing that the "actual answer" you've been given is wrong, you might want to consider a different approach to the problem -- integration by parts. You have an integral of the form 
$$
\int_0^1 f(x) g'(x) ~ dx
$$
(where $g(x) = ((1-x^2)x^5)'$), so you can convert it, with "parts", as follows:
\begin{align}
I &= \int_{0}^{1}4x^3\left(\dfrac{d^2}{dx^2}(1-x^2)x^5\right)dx\\
&= \int_{0}^{1}4x^3\left(\dfrac{d}{dx}\left(\frac{d}{dx}[(1-x^2)x^5]\right)\right)dx \\
&= \left. 4x^3 \left(\frac{d}{dx}[(1-x^2)x^5]\right)\right|_0^1 - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
\end{align}
In the left hand term, when $x = 0$ the $4x^3$ factor is zero; when $x = 1$, the "derivative" factor turns out to be $5 - 7 = -2$, so we have
\begin{align}
I 
&= 4 \cdot (-2) - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
&= -8 - \int_{0}^{1}12x^2\left(\frac{d}{dx}[(1-x^2)x^5]\right)dx \\
\end{align}
and we can repeat the integration by parts, to get 
\begin{align}
I &= -8 - \left.\left(12x^2 \cdot (1-x^2)x^5\right) \right|_0^1 +  \int_{0}^{1}24x\left((1-x^2)x^5\right)dx \\
\end{align}
In this case, the middle term is zero at both $x = 0$ and $x = 1$, so we have
\begin{align}
I 
&= -8 +  \int_{0}^{1}24x\left((1-x^2)x^5\right)dx \\
&= -8 +  \int_{0}^{1}24x(x^5-x^7)dx \\
&= -8 +  24\int_{0}^{1}(x^6-x^8)dx \\
&= -8 +  24 \left(\left. \frac{x^7}{7}-\frac{x^9}{9}\right|_{0}^{1}\right)\\
&= -8 +  24 \left( \frac{1}{7}-\frac{1}{9}\right)\\
&= -8 +  24 \left( \frac{2}{63}\right)\\
&= -8 +  \frac{16}{21}\\
\end{align}
Is this really the best way to go? Probably not -- I had to multiply stuff out in the middle to do the evaluations from $0$ to $1$, etc. -- but the general notion that whenever you see $fg'$ inside an integral, it's not a bad idea to try integration by parts, is still worth knowing. 
