Blog: `int (x^2 - 2x - 1)/((x - 1)^2(x^2 + 1)) dx` Evaluate the integral

Integrate $\displaystyle\int\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{{\left({x}-{1}\right)}}^{{2}}{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$

Rewrite the rational function using partial fractions.

$\displaystyle\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{{\left({x}-{1}\right)}}^{{2}}{\left({{x}}^{{2}}+{1}\right)}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{\left({x}-{1}\right)}}^{{2}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}+{1}}}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{1}\right)}+{B}{\left({{x}}^{{2}}+{1}\right)}+{\left({C}{x}+{D}\right)}{{\left({x}-{1}\right)}}^{{2}}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{\left({{x}}^{{3}}-{{x}}^{{2}}+{x}-{1}\right)}+{B}{{x}}^{{2}}+{B}+{\left({C}{x}+{D}\right)}{\left({{x}}^{{2}}-{2}{x}+{1}\right)}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{{x}}^{{3}}-{A}{{x}}^{{2}}-{A}{x}-{A}+{B}{{x}}^{{2}}+{B}$

$\displaystyle+{C}{{x}}^{{3}}-{2}{C}{{x}}^{{2}}+{C}{x}+{D}{{x}}^{{2}}-{2}{D}{x}+{D}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={\left({A}+{C}\right)}{{x}}^{{3}}+{\left(-{A}+{B}-{2}{C}+{D}\right)}{{x}}^{{2}}$

$\displaystyle+{\left({A}+{C}-{2}{D}\right)}{x}+{\left(-{A}+{B}+{D}\right)}$

Equate coefficients and solve for A, B, C, and D.

$\displaystyle{0}={A}+{C}$

$\displaystyle{A}=-{C}$

$\displaystyle-{2}={A}+{C}-{2}{D}$

$\displaystyle-{2}=-{C}+{C}-{2}{D}$

$\displaystyle-{2}=-{2}{D}$

$\displaystyle{D}={1}$

$\displaystyle{1}=-{A}+{B}-{2}{C}+{D}$

$\displaystyle{1}={C}+{B}-{2}{C}+{1}$

$\displaystyle{0}=-{1}{C}+{B}$

$\displaystyle{B}={C}$

$\displaystyle-{1}=-{A}+{B}+{D}$

$\displaystyle-{1}={C}+{B}+{1}$

$\displaystyle-{2}={B}+{B}$

$\displaystyle-{2}={2}{B}$

$\displaystyle{B}=-{1}$

$\displaystyle{C}=-{1}$

$\displaystyle{A}={1}$

$\displaystyle\int\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{{\left({x}-{1}\right)}}^{{2}}{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[\frac{{1}}{{{x}-{1}}}\right]}{\left.{d}{x}\right.}-\int{\left[\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}\right]}{\left.{d}{x}\right.}+\int{\left[\frac{{-{1}{x}+{1}}}{{{{x}}^{{2}}+{1}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[\frac{{1}}{{{x}-{1}}}\right]}{\left.{d}{x}\right.}-\int{\left[\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}\right]}{\left.{d}{x}\right.}+\int{\left[-\frac{{x}}{{{{x}}^{{2}}+{1}}}\right]}{\left.{d}{x}\right.}+\int{\left[\frac{{1}}{{{{x}}^{{2}}+{1}}}\right]}{\left.{d}{x}\right.}$

The first integral follows the pattern $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int{\left[\frac{{1}}{{{x}-{1}}}\right]}{\left.{d}{x}\right.}={\ln}{\left|{x}-{1}\right|}+{C}$

Integrate the second integral using u-substitution.

...

Integrate $\displaystyle\int\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{{\left({x}-{1}\right)}}^{{2}}{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$

Rewrite the rational function using partial fractions.

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{1}\right)}+{B}{\left({{x}}^{{2}}+{1}\right)}+{\left({C}{x}+{D}\right)}{{\left({x}-{1}\right)}}^{{2}}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{\left({{x}}^{{3}}-{{x}}^{{2}}+{x}-{1}\right)}+{B}{{x}}^{{2}}+{B}+{\left({C}{x}+{D}\right)}{\left({{x}}^{{2}}-{2}{x}+{1}\right)}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={A}{{x}}^{{3}}-{A}{{x}}^{{2}}-{A}{x}-{A}+{B}{{x}}^{{2}}+{B}$

$\displaystyle+{C}{{x}}^{{3}}-{2}{C}{{x}}^{{2}}+{C}{x}+{D}{{x}}^{{2}}-{2}{D}{x}+{D}$

$\displaystyle{{x}}^{{2}}-{2}{x}-{1}={\left({A}+{C}\right)}{{x}}^{{3}}+{\left(-{A}+{B}-{2}{C}+{D}\right)}{{x}}^{{2}}$

$\displaystyle+{\left({A}+{C}-{2}{D}\right)}{x}+{\left(-{A}+{B}+{D}\right)}$

Equate coefficients and solve for A, B, C, and D.

$\displaystyle{0}={A}+{C}$

$\displaystyle{A}=-{C}$

$\displaystyle-{2}={A}+{C}-{2}{D}$

$\displaystyle-{2}=-{C}+{C}-{2}{D}$

$\displaystyle-{2}=-{2}{D}$

$\displaystyle{D}={1}$

$\displaystyle{1}=-{A}+{B}-{2}{C}+{D}$

$\displaystyle{1}={C}+{B}-{2}{C}+{1}$

$\displaystyle{0}=-{1}{C}+{B}$

$\displaystyle{B}={C}$

$\displaystyle-{1}=-{A}+{B}+{D}$

$\displaystyle-{1}={C}+{B}+{1}$

$\displaystyle-{2}={B}+{B}$

$\displaystyle-{2}={2}{B}$

$\displaystyle{B}=-{1}$

$\displaystyle{C}=-{1}$

$\displaystyle{A}={1}$

$\displaystyle\int\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{{\left({x}-{1}\right)}}^{{2}}{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[\frac{{1}}{{{x}-{1}}}\right]}{\left.{d}{x}\right.}-\int{\left[\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}\right]}{\left.{d}{x}\right.}+\int{\left[-\frac{{x}}{{{{x}}^{{2}}+{1}}}\right]}{\left.{d}{x}\right.}+\int{\left[\frac{{1}}{{{{x}}^{{2}}+{1}}}\right]}{\left.{d}{x}\right.}$

The first integral follows the pattern $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int{\left[\frac{{1}}{{{x}-{1}}}\right]}{\left.{d}{x}\right.}={\ln}{\left|{x}-{1}\right|}+{C}$

Integrate the second integral using u-substitution.

Let $\displaystyle{u}={x}-{1}$

$\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={1}$

$\displaystyle{d}{u}={\left.{d}{x}\right.}$

$\displaystyle-\int\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}{\left.{d}{x}\right.}$

= $\displaystyle-\int{{u}}^{{-{{2}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{u}}+{C}$

$\displaystyle\frac{{1}}{{{x}-{1}}}+{C}$

Integrate the third integral using u-subsitution.

Let $\displaystyle{u}={{x}}^{{2}}+{1}$

$\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={2}{x}$

$\displaystyle{\left({\left.{d}{x}\right.}\right)}=\frac{{{d}{u}}}{{{2}{x}}}$

$\displaystyle-\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle=-\int\frac{{{x}}}{{{u}}}\cdot\frac{{{d}{u}}}{{{2}{x}}}$

$\displaystyle=-\frac{{1}}{{2}}{\ln}{\left|{u}\right|}+{C}$

$\displaystyle=-\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+{C}$

The fourth integral matches the pattern

$\displaystyle\int\frac{{\left.{d}{x}}}{{{{x}}^{{2}}+{{a}}^{{2}}}}={\left(\frac{{1}}{{a}}\right)}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{a}}\right)}+{C}$

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle={{\tan}}^{{-{{1}}}}{\left({x}\right)}+{C}$

The final answer is:

$\displaystyle{\ln}{\left|{x}-{1}\right|}+\frac{{1}}{{{x}-{1}}}-\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+{{\tan}}^{{-{{1}}}}{\left({x}\right)}+{C}$