New Notes Blog: `int (x^2+5) / (x^3-x^2+x+3) dx` Use partial fractions to find the indefinite integral

Friday, 14 August 2015

$\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}$ Use partial fractions to 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

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

To determine the indefinite integral of $\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}$ , we apply partial fraction decomposition to expand the integrand: $\displaystyle{f{{\left({x}\right)}}}=\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}$

The pattern on setting up partial fractions will depend on the factors of the denominator. The factored form of $\displaystyle{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}={\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}$ .

For the linear factor $\displaystyle{\left({x}+{1}\right)}$ , we will have partial fraction: $\displaystyle\frac{{A}}{{{x}+{1}}}$ .

For the quadratic factor $\displaystyle{\left({{x}}^{{2}}-{2}{x}+{3}\right)}$ , we will have partial fraction: $\displaystyle\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}-{2}{x}+{3}}}$ .

The integrand becomes:

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}=\frac{{A}}{{{x}+{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}-{2}{x}+{3}}}$

Multiply both side by the $\displaystyle{L}{C}{D}={\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}$ .

$\displaystyle{\left(\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}\right)}\cdot{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}={\left(\frac{{A}}{{{x}+{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}-{2}{x}+{3}}}\right)}\cdot{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}$

$\displaystyle{{x}}^{{2}}+{5}={A}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}+{\left({B}{x}+{C}\right)}{\left({x}+{1}\right)}$

We apply zero-factor property on $\displaystyle{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}$ to solve for values we can assign on x.

$\displaystyle{x}+{1}={0}$ then $\displaystyle{x}=-{1}$

$\displaystyle{{x}}^{{2}}-{2}{x}+{3}={0}{t}{h}{e}{n}{x}={1}\pm\sqrt{{{2}}}{i}$

To solve for $\displaystyle{A}$ , we plug-in $\displaystyle{x}=-{1}$ :

$\displaystyle{{\left(-{1}\right)}}^{{2}}+{5}={A}{\left({{\left(-{1}\right)}}^{{2}}-{2}\cdot{\left(-{1}\right)}+{3}\right)}+{\left({B}\cdot{\left(-{1}\right)}+{C}\right)}{\left(-{1}+{1}\right)}$

$\displaystyle{1}+{5}={A}{\left({1}+{2}+{3}\right)}+{\left(-{B}+{C}\right)}\cdot{0}$

$\displaystyle{6}={6}{A}$

$\displaystyle\frac{{6}}{{6}}=\frac{{{6}{A}}}{{6}}$

$\displaystyle{A}={1}$

To solve for $\displaystyle{C}$ , plug-in $\displaystyle{A}={1}$ and $\displaystyle{x}={0}$ so that $\displaystyle{B}\cdot{x}$ becomes $\displaystyle{0}$ :

$\displaystyle{{0}}^{{2}}+{5}={1}{\left({{0}}^{{2}}-{2}\cdot{0}+{3}\right)}+{\left({B}\cdot{0}+{C}\right)}{\left({0}+{1}\right)}$

$\displaystyle{0}+{5}={1}{\left({0}-{0}+{3}\right)}+{\left({0}+{C}\right)}{\left({1}\right)}$

$\displaystyle{5}={3}+{C}$

$\displaystyle{C}={5}-{3}$

$\displaystyle{C}={2}$ .

To solve for $\displaystyle{B}$ , plug-in $\displaystyle{A}={1}$ , $\displaystyle{C}={2}$ , and $\displaystyle{x}={1}$ :

$\displaystyle{{1}}^{{2}}+{5}={1}{\left({{1}}^{{2}}-{2}\cdot{1}+{3}\right)}+{\left({B}\cdot{1}+{2}\right)}{\left({1}+{1}\right)}$

$\displaystyle{1}+{5}={1}{\left({1}-{2}+{3}\right)}+{\left({B}+{2}\right)}{\left({2}\right)}$

$\displaystyle{6}={2}+{2}{B}+{4}$

$\displaystyle{2}{B}={6}-{2}-{4}$

$\displaystyle{2}{B}={0}$

$\displaystyle\frac{{{2}{B}}}{{2}}=\frac{{0}}{{2}}$

$\displaystyle{B}={0}$

Plug-in $\displaystyle{A}={1}$ , $\displaystyle{B}={0}$ , and $\displaystyle{C}={2}$ , we get the partial fraction decomposition:

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}=\frac{{1}}{{{x}+{1}}}+\frac{{{0}{x}+{2}}}{{{{x}}^{{2}}-{2}{x}+{3}}}$

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

The integral becomes:

$\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}=\int{\left[\frac{{1}}{{{x}+{1}}}+\frac{{2}}{{{{x}}^{{2}}-{2}{x}+{3}}}\right]}{\left.{d}{x}\right.}$

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

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

For the first integral, we apply integration formula for logarithm: $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}$ .

Let $\displaystyle{u}={x}+{1}$ then $\displaystyle{d}{u}={\left.{d}{x}\right.}$

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

$\displaystyle={\ln}{\left|{u}\right|}$

$\displaystyle={\ln}{\left|{x}+{1}\right|}$

Apply indefinite integration formula for rational function:

$\displaystyle\int\frac{{1}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}{\left.{d}{x}\right.}=\frac{{2}}{\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}{\arctan{{\left(\frac{{{2}{a}{x}+{b}}}{\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}\right)}}}+{C}$

By comparing " $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ " with " $\displaystyle{{x}}^{{2}}-{2}{x}+{3}$ ", we determine the corresponding values: $\displaystyle{a}={1}$ , $\displaystyle{b}=-{2}$ , and $\displaystyle{c}={3}$ .

The second integral becomes:

$\displaystyle\int\frac{{2}}{{{{x}}^{{2}}-{2}{x}+{3}}}{\left.{d}{x}\right.}={2}\int\frac{{1}}{{{{x}}^{{2}}-{2}{x}+{3}}}{\left.{d}{x}\right.}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{\sqrt{{{4}\cdot{1}\cdot{3}-{{\left(-{2}\right)}}^{{2}}}}}{\arctan{{\left(\frac{{{2}\cdot{1}{x}+{\left(-{2}\right)}}}{\sqrt{{{4}\cdot{1}\cdot{3}-{{\left(-{2}\right)}}^{{2}}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{\sqrt{{{12}-{4}}}}{\arctan{{\left(\frac{{{2}{x}-{2}}}{\sqrt{{{12}-{4}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{\sqrt{{{8}}}}{\arctan{{\left(\frac{{{2}{x}-{2}}}{\sqrt{{{8}}}}\right)}}}\right]}$

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

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

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

Combining the results, we get the indefinite integral as:

$\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}={\ln}{\left|{x}+{1}\right|}+\frac{{{2}{\arctan{{\left(\frac{{{x}-{1}}}{\sqrt{{{2}}}}\right)}}}}}{\sqrt{{{2}}}}+{C}$

$\displaystyle={\ln}{\left|{x}+{1}\right|}+\sqrt{{{2}}}{\arctan{{\left(\frac{{\sqrt{{{2}}}{\left({x}-{1}\right)}}}{{2}}\right)}}}+{C}$

$\displaystyle={\ln}{\left|{x}+{1}\right|}+\sqrt{{{2}}}{\arctan{{\left(\frac{{{x}\sqrt{{{2}}}-\sqrt{{{2}}}}}{{2}}\right)}}}+{C}$

New Notes Blog

Friday, 14 August 2015

$\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Friday, 14 August 2015

Use partial fractions to find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{{{x}}^{{2}}+{5}}}{{{{x}}^{{3}}-{{x}}^{{2}}+{x}+{3}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?