Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)
F(1) = 1/3
F(0) = 0
= 1/3
If you think of it as $\displaystyle \int af\prime(x)~f(x)^n~=~\frac{af(x)^{n+1}}{n+1}$
Then $\displaystyle f(x)=4x^3+2 \implies f\prime(x)=12x^2$
Hence a must be $\displaystyle \frac{1}{12}$
Given the general result we have $\displaystyle \frac{\frac{1}{12}(4x^3+2)^4}{4}$
$\displaystyle
=~\frac{1}{48}(4x^3+2)^4
$
Probably you have seen what I've done. Only in case you need some confirmation I'm going to show you how I got the anit-derivative:
$\displaystyle \int_0^1\left(x^2 (4x^3+2)^3\right) dx = \int_0^1\left(x^2\cdot 2^3 \cdot (2x^3+1)^3\right) dx$
Let $\displaystyle u = 2x^3+1~\implies~\frac{du}{dx}=6x^2~\implies~du = 6x^2 dx$
That means:
$\displaystyle du = \frac43 \cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2} \cdot dx$
The integral becomes:
$\displaystyle \int_0^1\left(\frac43 \underbrace{\cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2}}_{\frac{du}{dx}} \cdot (\underbrace{2x^3+1}_{u})^3)\right) dx = \frac43 \cdot \int_{u(0)}^{u(1)} (u^3) du$ = $\displaystyle \frac43 \cdot \left[\frac14 \cdot u^4\right]_{u(0)}^{u(1)}$
Now re-substitute and you'll get:
$\displaystyle
\int_0^1\left(x^2 (4x^3+2)^3\right) dx =\left. \frac{(2x^3+1)^4}{3}\right|_0^1$
Introduce a new variable:
u = 4x^3 + 2
They you can get dx=du/12x^2. You insert both of these two into you initial equation and get:
x^2*u^3*du/12x^2
Due to new variable you must change limits into 2 to 6, getting final equation
u^4/48Inserting the limits you obtain the result 26.67 or 80/3.