Step 1: Use the cover-up method:
At $x=1$ the factor $(1-x)$ vanishes: residue $\dfrac{2(1)}{4+1}=\dfrac25$. At $x=-4$ the factor $(4+x)$ vanishes: residue $\dfrac{2(-4)}{1-(-4)}=-\dfrac85$.
Step 2: Write the decomposition:
$\dfrac{2x}{4-3x-x^2}=\dfrac{2/5}{1-x}-\dfrac{8/5}{4+x}$.
Step 3: Integrate term by term:
$\int\dfrac{dx}{1-x}=-\ln|1-x|$, so the integral is $-\dfrac25\ln|1-x|-\dfrac85\ln|4+x|+c$. Option D.
Final Answer:
Partial fractions give coefficients 2/5 and -8/5.
\[ \boxed{\text{(D) }-\dfrac85\log|4+x|-\dfrac25\log|1-x|+c} \]