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

Friday, April 6, 2012

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

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

Let's evaluate the above integral by rewriting the integrand as,

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

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

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

Now again rewrite the second integral,

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

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

Now let's evaluate the above three integrals,

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

Let's use the integral substitution,

Let u=x-1,

du=dx

$\displaystyle=\int\frac{{1}}{{{1}+{{u}}^{{2}}}}{d}{u}$

The above can be evaluated using the standard integral,

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

$\displaystyle=\frac{{1}}{{1}}{{\tan}}^{{-{1}}}{\left(\frac{{u}}{{1}}\right)}$

$\displaystyle={{\tan}}^{{-{1}}}{\left({u}\right)}$

Substitute back u=x-1,

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

Now let's evaluate the second integral by integral substitution,

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

Let $\displaystyle{v}={\left({{x}}^{{2}}-{2}{x}+{2}\right)}$

$\displaystyle{d}{v}={\left({2}{x}-{2}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{v}}^{{2}}}{d}{v}$

$\displaystyle=\frac{{{v}}^{{-{2}+{1}}}}{{-{2}+{1}}}$

$\displaystyle=\frac{{{v}}^{{-{1}}}}{{-{{1}}}}$

$\displaystyle=-\frac{{1}}{{v}}$

Substitute back $\displaystyle{v}={\left({{x}}^{{2}}-{2}{x}+{2}\right)}$

$\displaystyle=-\frac{{1}}{{{{x}}^{{2}}-{2}{x}+{2}}}$

Noe let's evaluate the third integral,

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

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

Let's use the integral substitution,

Let $\displaystyle{\tan{{\left({y}\right)}}}={x}-{1}$

$\displaystyle{{\sec}}^{{2}}{\left({y}\right)}{\left.{d}{y}\right.}={\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{{{\sec}}^{{2}}{\left({y}\right)}{\left.{d}{y}\right.}}}{{{\left({{\tan}}^{{2}}{\left({y}\right)}+{1}\right)}}^{{2}}}$

Using the identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left({y}\right)}={{\sec}}^{{2}}{\left({y}\right)}$

$\displaystyle=\int\frac{{{{\sec}}^{{2}}{\left({y}\right)}}}{{{\left({{\sec}}^{{2}}{\left({y}\right)}\right)}}^{{2}}}{\left.{d}{y}\right.}$

$\displaystyle=\int\frac{{1}}{{{{\sec}}^{{2}}{\left({y}\right)}}}{\left.{d}{y}\right.}$

$\displaystyle=\int{{\cos}}^{{2}}{\left({y}\right)}{\left.{d}{y}\right.}$

Now use the identity: $\displaystyle{{\cos}}^{{2}}{\left({y}\right)}=\frac{{{1}+{\cos{{\left({2}{y}\right)}}}}}{{2}}$

$\displaystyle=\int\frac{{{1}+{\cos{{\left({2}{y}\right)}}}}}{{2}}{\left.{d}{y}\right.}$

$\displaystyle=\int\frac{{{1}{\left.{d}{y}\right.}}}{{2}}+\int\frac{{\cos{{\left({2}{y}\right)}}}}{{2}}{\left.{d}{y}\right.}$

$\displaystyle=\frac{{y}}{{2}}+\frac{{1}}{{2}}\frac{{\sin{{\left({2}{y}\right)}}}}{{2}}$

$\displaystyle=\frac{{y}}{{2}}+\frac{{\sin{{\left({2}{y}\right)}}}}{{4}}$

Substitute back $\displaystyle{y}={{\tan}}^{{-{1}}}{\left({x}-{1}\right)}$

$\displaystyle=\frac{{1}}{{2}}{\arctan{{\left({x}-{1}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{\arctan{{\left({x}-{1}\right)}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\arctan{{\left({x}-{1}\right)}}}+\frac{{1}}{{4}}{\left\lbrace{2}{\sin{{\left({\arctan{{\left({x}-{1}\right)}}}\right)}}}{\cos{{\left({\arctan{{\left({x}-{1}\right)}}}\right)}}}\right\rbrace}$

$=1/2arctan(x-1)+1/4{2*(x-1)/sqrt(x^2-2x+2)*1/sqrt(x^2-2x+2)}$

$\displaystyle=\frac{{1}}{{2}}{\arctan{{\left({x}-{1}\right)}}}+{\left(\frac{{1}}{{2}}\right)}\frac{{{x}-{1}}}{{{{x}}^{{2}}-{2}{x}+{2}}}$

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