(2) f(g(x)) cdot f(u)=(u-1)^3+2 u^2, g(x)=x+1

To solve the given integral, we need to find the antiderivative of the function $(g(x))g(u)$ with respect to $u$. <br /><br />Given that $g(x) = x + 1$, we can substitute this into the integral:<br /><br />$\int (g(x))g(u) du = \int (u + 1)(u - 1)^3 + 2u^2 du$<br /><br />Expanding the expression inside the integral, we have:<br /><br />$\int (u^2 - u)(u - 1)^3 + 2u^2 du$<br /><br />Expanding further, we get:<br /><br />$\int (u^2 - u)(u^3 - 3u^2 + 3u - 1) + 2u^2 du$<br /><br />Expanding and simplifying, we have:<br /><br />$\int (u^5 - u^4 + 3u^3 - 4u^2 + 3u) + 2u^2 du$<br /><br />Integrating term by term, we get:<br /><br />$\frac{1}{6}u^6 - \frac{1}{5}u^5 + \frac{3}{4}u^4 - \frac{4}{3}u^3 + \frac{3}{2}u^2 + 2u^2 + C$<br /><br />Simplifying further, we have:<br /><br />$\frac{1}{6}u^6 - \frac{1}{5}u^5 + \frac{3}{4}u^4 - \frac{4}{3}u^3 + \frac{7}{2}u^2 + C$<br /><br />Therefore, the correct answer is $\frac{1}{6}u^6 - \frac{1}{5}u^5 + \frac{3}{4}u^4 - \frac{4}{3}u^3 + \frac{7}{2}u^2 + C$.