New Notes Blog: `int (x^3+3x-4)/(x^3-4x^2+4x) dx` Use partial fractions to find the indefinite integral

Friday, 20 June 2014

$\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

For the given integral problem: $\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}$ , we may simplify by applying long division since the highest degree of x is the same from numerator and denominator side.

$\displaystyle\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}={1}+\frac{{{4}{{x}}^{{2}}-{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}$ .

Apply partial fraction decomposition on the expression $\displaystyle\frac{{{4}{{x}}^{{2}}-{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}$ .

The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the factored form of the denominator will be:

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

$\displaystyle={\left({x}\right)}{\left({x}-{2}\right)}{\left({x}-{2}\right)}$ or $\displaystyle{x}{{\left({x}-{2}\right)}}^{{2}}$

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

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

The rational expression becomes:

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

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

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

$\displaystyle{4}{{x}}^{{2}}-{x}-{4}={A}\cdot{{\left({x}-{2}\right)}}^{{2}}+{B}\cdot{\left({x}{\left({x}-{2}\right)}\right)}+{C}\cdot{x}$

We apply zero-factor property on x(x-2)^2 to solve for value we can assign on x.

$\displaystyle{x}={0}$

$\displaystyle{x}-{2}={0}$ then $\displaystyle{x}={2}$ .

To solve for $\displaystyle{A}$ , we plug-in $\displaystyle{x}={0}$ :

$\displaystyle{4}\cdot{{0}}^{{2}}-{0}-{4}={A}\cdot{{\left({0}-{2}\right)}}^{{2}}+{B}\cdot{\left({0}{\left({0}-{2}\right)}\right)}+{C}\cdot{0}$

$\displaystyle{0}-{0}-{4}={A}\cdot{{\left(-{2}\right)}}^{{2}}+{0}+{0}$

$\displaystyle-{4}={4}{A}$

$\displaystyle-\frac{{4}}{{4}}=\frac{{{4}{A}}}{{4}}$

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

To solve for $\displaystyle{C}$ , we plug-in $\displaystyle{x}={2}$ :

$\displaystyle{4}\cdot{{2}}^{{2}}-{2}-{4}={A}\cdot{{\left({2}-{2}\right)}}^{{2}}+{B}\cdot{\left({2}{\left({2}-{2}\right)}\right)}+{C}\cdot{2}$

$\displaystyle{16}-{2}-{4}={A}\cdot{0}+{B}\cdot{0}+{2}{C}$

$\displaystyle{10}={0}+{0}+{2}{C}$

$\displaystyle{10}={2}{C}$

$\displaystyle\frac{{{10}}}{{2}}=\frac{{{2}{C}}}{{2}}$

$\displaystyle{C}={5}$

To solve for B, plug-in $\displaystyle{x}={1}$ , $\displaystyle{A}=-{1}$ , and $\displaystyle{C}={5}$ :

$\displaystyle{4}\cdot{{1}}^{{2}}-{1}-{4}={\left(-{1}\right)}\cdot{{\left({1}-{2}\right)}}^{{2}}+{B}\cdot{\left({1}{\left({1}-{2}\right)}\right)}+{5}\cdot{1}$

$\displaystyle{4}-{1}-{4}={\left(-{1}\right)}\cdot{{\left(-{1}\right)}}^{{2}}+{B}{\left({1}\cdot{\left(-{1}\right)}\right)}+{5}$

$\displaystyle-{1}=-{1}-{B}+{5}$

$\displaystyle-{1}=-{B}+{4}$

$\displaystyle-{1}-{4}=-{B}$

$\displaystyle-{5}=-{B}$

$\displaystyle\frac{{-{5}}}{{-{1}}}=\frac{{-{B}}}{{-{1}}}$

$\displaystyle{B}={5}$

Plug-in $\displaystyle{A}=-{1}$ , $\displaystyle{B}={5},$ and $\displaystyle{C}={5}$ , we get the partial decomposition:

$\displaystyle\frac{{{4}{{x}}^{{2}}-{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}=-\frac{{1}}{{x}}+\frac{{5}}{{{x}-{2}}}+\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}$

Then the integrand becomes:

$\displaystyle\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}={1}+\frac{{{4}{{x}}^{{2}}-{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}$ .

$\displaystyle={1}-\frac{{1}}{{x}}+\frac{{5}}{{{x}-{2}}}+\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}$

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

$\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}=\int{\left[{1}-\frac{{1}}{{x}}+\frac{{5}}{{{x}-{2}}}+\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{1}{\left.{d}{x}\right.}-\int\frac{{1}}{{x}}{\left.{d}{x}\right.}+\int\frac{{5}}{{{x}-{2}}}{\left.{d}{x}\right.}+\int\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}{\left.{d}{x}\right.}$

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

$\displaystyle\int{1}{\left.{d}{x}\right.}={1}{x}$ or $\displaystyle{x}$

Apply integration formula for logarithm: $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}$ .

$\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

$\displaystyle\int\frac{{5}}{{{x}-{2}}}{\left.{d}{x}\right.}=\int\frac{{5}}{{u}}{d}{u}$

$\displaystyle={5}{\ln}{\left|{u}\right|}$

$\displaystyle={5}{\ln}{\left|{x}-{2}\right|}$

Note: Let $\displaystyle{u}={x}-{2}$ then $\displaystyle{d}{u}={\left.{d}{x}\right.}$ .

Apply the Power Rule for integration: $\displaystyle\int{\left({{u}}^{{n}}\right)}{\left.{d}{x}\right.}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle\int\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}{\left.{d}{x}\right.}=\int\frac{{5}}{{{u}}^{{2}}}{d}{u}$

$\displaystyle=\int{5}{{u}}^{{-{2}}}{d}{u}$

$\displaystyle={5}\cdot\frac{{{u}}^{{-{2}+{1}}}}{{-{2}+{1}}}$

$\displaystyle={5}\cdot\frac{{{u}}^{{-{{1}}}}}{{-{1}}}$

$\displaystyle=-\frac{{5}}{{u}}$

$\displaystyle=-\frac{{5}}{{{x}-{2}}}$

Note: Let $\displaystyle{u}={x}-{2}$ then $\displaystyle{d}{u}={\left.{d}{x}\right.}$

Combining the results, we get the indefinite integral as:

$\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}={x}-{\ln}{\left|{x}\right|}+{5}{\ln}{\left|{x}-{2}\right|}-\frac{{5}}{{{x}-{2}}}+{C}$

New Notes Blog

Friday, 20 June 2014

$\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}$ 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, 20 June 2014

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?

$\displaystyle\int\frac{{{{x}}^{{3}}+{3}{x}-{4}}}{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}}}{\left.{d}{x}\right.}$ 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?