Answer
$ \displaystyle \frac{e^{(x-1)^{3}}}{3}+C$
Work Step by Step
see Substitution RuIe, p.962:
1. Write $u$ a{\it s} a function of x.
2. Take the derivative $du/dx$ and solve for the quantity $dx$ in terms of $du$.
3. Use the expression you obtain in step 2 to substitute for $dx$ in the given integral and substitute $u$ for its defining expression.
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(1) Given $\quad u=(x-1)^{3}$
(2) (chain rule) $\quad du=3(x-1)^{2}\cdot 1dx\ \ \Rightarrow$
$ \displaystyle \ \ (x-1)^{2}dx=\frac{du}{3}$
$(dx=\displaystyle \frac{du}{3(x-1)^{2}})$
$(3)$
$\displaystyle \int(x-1)^{2}e^{(x-1)^{3}}dx$ = $\displaystyle \int e^{(x+1)^{3}}[(x-1)^{2}dx]$
$=\displaystyle \int e^{u}(\frac{du}{3})$= ... constant multiple
$=\displaystyle \frac{1}{3}\int e^{u}du$
$=\displaystyle \frac{e^{u}}{3}+C$= ... bring back x
$= \displaystyle \frac{e^{(x-1)^{3}}}{3}+C$
