New Notes Blog: `int (x^3+x+1)/(x^4+2x^2+1) dx` Find the indefinite integral

Sunday, 30 August 2015

$\displaystyle\int\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Indefinite integral are written in the form of $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

To determine the indefinite integral of $\displaystyle\int\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ , we apply partial fraction decomposition to expand the integrand: $\displaystyle{f{{\left({x}\right)}}}=\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}$ .

The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the denominator is in a similar form of perfect squares trinomial: $\displaystyle{{x}}^{{2}}+{2}{x}{y}+{{y}}^{{2}}={{\left({x}+{y}\right)}}^{{2}}$

Applying the special factoring on $\displaystyle{\left({{x}}^{{4}}+{2}{{x}}^{{2}}+{1}\right)}$ , we get: $\displaystyle{\left({{x}}^{{4}}+{2}{x}+{1}\right)}={{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$ .

For the repeated quadratic factor $\displaystyle{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$ , we will have partial fraction: $\displaystyle\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{1}}}+\frac{{{C}{x}+{D}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}$ .

The integrand becomes:

$\displaystyle\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}=\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{1}}}+\frac{{{C}{x}+{D}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}$

Multiply both sides by the $\displaystyle{L}{C}{D}={{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$ :

$\displaystyle{\left(\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}\right)}\cdot{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}={\left(\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{1}}}+\frac{{{C}{x}+{D}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}\right)}\cdot{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$

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

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

$\displaystyle{{x}}^{{3}}+{0}{{x}}^{{2}}+{x}+{1}={A}{{x}}^{{3}}+{A}{x}+{B}{{x}}^{{2}}+{B}+{C}{x}+{D}$

Equate the coefficients of similar terms on both sides to list a system of equations:

Terms with $\displaystyle{{x}}^{{3}}$ : $\displaystyle{1}={A}$

Terms with $\displaystyle{{x}}^{{2}}$ : $\displaystyle{0}={B}$

Terms with $\displaystyle{x}$ : $\displaystyle{1}={A}+{C}$

Plug-in $\displaystyle{A}={1}$ on $\displaystyle{1}={A}+{C}$ , we get:

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

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

$\displaystyle{C}={0}$

Constant terms: $\displaystyle{1}={B}+{D}$

Plug-in $\displaystyle{B}={0}$ on $\displaystyle{1}={B}+{D}$ , we get:

$\displaystyle{1}={0}+{D}$

$\displaystyle{D}={1}$

Plug-in the values of $\displaystyle{A}={1}$ , $\displaystyle{B}={0}$ , $\displaystyle{C}={0}$ , and $\displaystyle{D}={1}$ , we get the partial fraction decomposition:

$\displaystyle\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}=\frac{{{1}{x}+{0}}}{{{{x}}^{{2}}+{1}}}+\frac{{{0}{x}+{1}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}$

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

Then the integral becomes:

$\displaystyle\int\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\int{\left[\frac{{x}}{{{{x}}^{{2}}+{1}}}+\frac{{1}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}\right]}{\left.{d}{x}\right.}$

Apply the basic integration property: $\displaystyle\int{\left({u}+{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}+\int{\left({v}\right)}{\left.{d}{x}\right.}.$

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

For the first integral, we apply integration formula for rational function as:

$\displaystyle\int\frac{{u}}{{{{u}}^{{2}}+{{a}}^{{2}}}}{d}{u}=\frac{{1}}{{2}}{\ln}{\left|{{u}}^{{2}}+{{a}}^{{2}}\right|}+{C}$

Then, $\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+{C}{\quad\text{or}\quad}\frac{{{\ln}{\left|{{x}}^{{2}}+{1}\right|}}}{{2}}+{C}$

For the second integral, we apply integration by trigonometric substitution.

We let $\displaystyle{x}={\tan{{\left({u}\right)}}}$ then $\displaystyle{\left.{d}{x}\right.}={{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

Plug-in the values, we get:

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

Apply the trigonometric identity: $\displaystyle{{\tan}}^{{2}}{\left({u}\right)}+{1}={{\sec}}^{{2}}{\left({u}\right)}$ and trigonometric property: $\displaystyle\frac{{1}}{{{{\sec}}^{{2}}{\left({u}\right)}}}={{\cos}}^{{2}}{\left({u}\right)}$

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

$\displaystyle=\int\frac{{1}}{{{{\sec}}^{{4}}{\left({u}\right)}}}\cdot{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

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

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

Apply the integration formula for cosine function: $\displaystyle\int{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\left[{x}+{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}\right]}+{C}$

$\displaystyle\int{{\cos}}^{{2}}{\left({u}\right)}{d}{u}=\frac{{1}}{{2}}{\left[{u}+{\sin{{\left({u}\right)}}}{\cos{{\left({u}\right)}}}\right]}+{C}$

Based from $\displaystyle{x}={\tan{{\left({u}\right)}}}$ then :

$\displaystyle{u}={\arctan{{\left({x}\right)}}}$

$\displaystyle{\sin{{\left({u}\right)}}}=\frac{{x}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

$\displaystyle{\cos{{\left({u}\right)}}}=\frac{{1}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

Then the integral becomes:

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

$\displaystyle=\frac{{1}}{{2}}{\left[{\arctan{{\left({x}\right)}}}+{\left(\frac{{x}}{\sqrt{{{{x}}^{{2}}+{1}}}}\right)}\cdot{\left(\frac{{1}}{\sqrt{{{{x}}^{{2}}+{1}}}}\right)}\right]}$

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

Combining the results, we get:

$\displaystyle\int\frac{{{{x}}^{{3}}+{x}+{1}}}{{{{x}}^{{4}}+{2}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\frac{{{\ln}{\left|{{x}}^{{2}}+{1}\right|}}}{{2}}+\frac{{\arctan{{\left({x}\right)}}}}{{2}}+\frac{{x}}{{{2}{{x}}^{{2}}+{2}}}+{C}$