Step 1: Pick the substitution.
For $I=\displaystyle\int\frac{\cos^3x}{(1+\sin x)^4}\,dx$, try $u=1+\sin x$ since $du=\cos x\,dx$ is already present.
Step 2: Split off one cosine.
Write $\cos^3x=\cos^2x\cos x=(1-\sin^2x)\cos x$, and $1-\sin^2x=(1-\sin x)(1+\sin x)$, so
\[ I=\int\frac{(1-\sin x)(1+\sin x)\cos x}{(1+\sin x)^4}\,dx=\int\frac{(1-\sin x)\cos x}{(1+\sin x)^3}\,dx. \]
Step 3: Substitute $u=1+\sin x$.
Then $\sin x=u-1$, so $1-\sin x=2-u$, and
\[ I=\int\frac{2-u}{u^3}\,du=\int\left(\frac{2}{u^3}-\frac{1}{u^2}\right)du. \]
Step 4: Integrate term by term.
\[ I=2\cdot\frac{u^{-2}}{-2}-\frac{u^{-1}}{-1}+c=-\frac{1}{u^2}+\frac{1}{u}+c. \]
Step 5: Combine into one fraction.
\[ I=\frac{-1+u}{u^2}+c=\frac{u-1}{u^2}+c=\frac{\sin x}{(1+\sin x)^2}+c. \]
Step 6: Match the given option.
Using $\cos^2x=(1-\sin x)(1+\sin x)$, this same antiderivative (up to a constant) can be written as $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$, which is the listed form.
\[ \boxed{-\dfrac{\cos^4x}{4(1+\sin x)^4}+c} \]