Answer
$$\frac{1}{9} \ln |x+1|-\frac{1}{9} \ln |x-2|-\frac{1}{3} \frac{1}{x-2}+C$$
Work Step by Step
Given $$\int \frac{d x}{x^{3}-3 x^{2}+4}$$
Since
\begin{aligned}
\frac{1}{x^{3}-3 x+4} &=\frac{1}{(x+1)(x-2)^{2}} \\
&=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{(x-2)^{2}} \\
&=\frac{(x-2)^{2} \mathrm{A}+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)}{x^{3}-3 x+4} \\
1 &=(x-2)^{2} \mathrm{A}+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)
\end{aligned}
Then
\begin{align*}
\text{at } x&=-1\ \ \ \ \to A= \frac{1}{9} \\
\text{at } x&=2\ \ \ \ \to B=\frac{-1}{9}\\
\text{at } x&=0\ \ \ \ \to C=\frac{1}{3}
\end{align*}
Hence
\begin{aligned}
\int \frac{d x}{x^{3}-3 x+4} &=\frac{1}{9} \int \frac{1}{x+1} d x-\frac{1}{9} \int \frac{1}{x-2} d x+\frac{1}{3} \int \frac{1}{(x-2)^{2}} d x\\
&=\frac{1}{9} \ln |x+1|-\frac{1}{9} \ln |x-2|-\frac{1}{3} \frac{1}{x-2}+C
\end{aligned}
