Question

$\int\frac{1x^{2}-2x - 1}{(x - 1)^{2}(x^{2}+1)}dx$

Explanation:

Step1: Use partial - fraction decomposition

Let $\frac{x^{2}-2x - 1}{(x - 1)^{2}(x^{2}+1)}=\frac{A}{x - 1}+\frac{B}{(x - 1)^{2}}+\frac{Cx+D}{x^{2}+1}$. Then $x^{2}-2x - 1=A(x - 1)(x^{2}+1)+B(x^{2}+1)+(Cx + D)(x - 1)^{2}$.
If $x = 1$, we have $1-2 - 1=B(1 + 1)$, so $B=-1$.
Expand the right - hand side:
\[

$$\begin{align*} A(x - 1)(x^{2}+1)+B(x^{2}+1)+(Cx + D)(x - 1)^{2}&=A(x^{3}+x-x^{2}-1)+B(x^{2}+1)+(Cx + D)(x^{2}-2x + 1)\\ &=A(x^{3}-x^{2}+x - 1)+Bx^{2}+B+(Cx^{3}-2Cx^{2}+Cx+Dx^{2}-2Dx + D)\\ &=(A + C)x^{3}+(-A + B-2C+D)x^{2}+(A + C-2D)x+(-A + B + D) \end{align*}$$

\]
Comparing the coefficients of $x^{3}$: $A + C=0$.
Comparing the coefficients of $x^{2}$: $-A + B-2C+D = 1$. Substitute $B=-1$, we get $-A-1-2C+D = 1$.
Comparing the coefficients of $x$: $A + C-2D=-2$.
Since $A + C = 0$, then $-2D=-2$, so $D = 1$.
Substitute $D = 1$ and $B=-1$ into $-A-1-2C+D = 1$, we have $-A-2C=1$. And because $A=-C$, then $C=-1$ and $A = 1$.
So $\frac{x^{2}-2x - 1}{(x - 1)^{2}(x^{2}+1)}=\frac{1}{x - 1}-\frac{1}{(x - 1)^{2}}-\frac{x - 1}{x^{2}+1}$.

Step2: Integrate term - by - term

\[

$$\begin{align*} \int\frac{x^{2}-2x - 1}{(x - 1)^{2}(x^{2}+1)}dx&=\int\frac{1}{x - 1}dx-\int\frac{1}{(x - 1)^{2}}dx-\int\frac{x}{x^{2}+1}dx+\int\frac{1}{x^{2}+1}dx\\ &=\ln|x - 1|+\frac{1}{x - 1}-\frac{1}{2}\ln(x^{2}+1)+\arctan(x)+C \end{align*}$$

\]

Answer:

$\ln|x - 1|+\frac{1}{x - 1}-\frac{1}{2}\ln(x^{2}+1)+\arctan(x)+C$