Answer
$$2 \ln |x-1|+\frac{1}{2} \ln \left|x^{2}+1\right|-3 \tan ^{-1} x+C$$
Work Step by Step
Given $$\int \frac{\left(3 x^{2}-4 x+5\right) d x}{(x-1)\left(x^{2}+1\right)}$$
Since
\begin{align*}
\frac{1}{(x-1)\left(x^{2}+1\right)}&=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+1}\\
&=\frac{A(x^{2}+1)+B(x-1)}{(x-1)\left(x^{2}+1\right)}\\
1&=A(x^{2}+1)+B(x-1)
\end{align*}
\begin{align*}
\text{at } x&=1 \ \ \ \ \to A=2 \\
\text{at } x&= 0\ \ \ \ \to C=-3\\
\text{at } x&= -1\ \ \ \ \to B =-1\\
\end{align*}
Hence $B=-1,\ \ \ C=0$ and
\begin{aligned}
\int \frac{3 x^{2}-4 x+5}{(x-1)\left(x^{2}+1\right)} d x &=\int \frac{2}{x-1} d x+\int \frac{x-3}{x^{2}+1} d x \\
&=\int \frac{2}{x-1} d x+\int \frac{\frac{1}{2} \cdot 2 x}{x^{2}+1} d x-\int \frac{3}{x^{2}+1} d x \\
&=2 \ln |x-1|+\frac{1}{2} \ln \left|x^{2}+1\right|-3 \tan ^{-1} x+C
\end{aligned}
