Step 1: Spot the pattern.
We want $\int\dfrac{x-1}{(x+1)^3}e^x\,dx$. Whenever an integral looks like $\int e^x[f(x)+f'(x)]\,dx$, the answer is simply $e^x f(x)+c$. Our plan is to write the bracket in that form.
Step 2: Split the numerator.
Write $x-1=(x+1)-2$. Then \[ \frac{x-1}{(x+1)^3}=\frac{(x+1)-2}{(x+1)^3}=\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}. \]
Step 3: Guess the function $f$.
Let $f(x)=\dfrac{1}{(x+1)^2}$. This is a sensible guess because differentiating it brings down a power.
Step 4: Differentiate the guess.
\[ f'(x)=-2(x+1)^{-3}=-\frac{2}{(x+1)^3}. \]
Step 5: Check the match.
Now $f(x)+f'(x)=\dfrac{1}{(x+1)^2}-\dfrac{2}{(x+1)^3}$, which is exactly our integrand bracket. So the integral fits the $e^x[f+f']$ form perfectly.
Step 6: Write the result.
Therefore \[ \int e^x\left[f(x)+f'(x)\right]dx=e^x f(x)+c=\frac{e^x}{(x+1)^2}+c. \]
\[ \boxed{\dfrac{e^x}{(x+1)^2}+c} \]