New Notes Blog: `int (2x^3-4x^2-15x+5)/(x^2-2x-8) dx` Use partial fractions to find the indefinite integral

Wednesday, 25 February 2015

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}$

The integrand is a improper rational function,as the degree of the numerator is greater than the degree of the denominator.So we have to carry out division.

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}={2}{x}+\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$

Since the polynomials do not completely divide, we have to continue with the partial fractions on the remainder and factor out the denominator.

$\displaystyle\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}=\frac{{{x}+{5}}}{{{{x}}^{{2}}-{4}{x}+{2}{x}-{8}}}$

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

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

Now we create the partial fraction template,

$\displaystyle\frac{{{x}+{5}}}{{{\left({x}-{4}\right)}{\left({x}+{2}\right)}}}=\frac{{A}}{{{x}-{4}}}+\frac{{B}}{{{x}+{2}}}$

Multiply the above equation with the denominator,

$\displaystyle\Rightarrow{\left({x}+{5}\right)}={A}{\left({x}+{2}\right)}+{B}{\left({x}-{4}\right)}$

$\displaystyle\Rightarrow{x}+{5}={A}{x}+{2}{A}+{B}{x}-{4}{B}$

$\displaystyle\Rightarrow{x}+{5}={\left({A}+{B}\right)}{x}+{2}{A}-{4}{B}$

Equating the...

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}$

The integrand is a improper rational function,as the degree of the numerator is greater than the degree of the denominator.So we have to carry out division.

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}={2}{x}+\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$

Since the polynomials do not completely divide, we have to continue with the partial fractions on the remainder and factor out the denominator.

$\displaystyle\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}=\frac{{{x}+{5}}}{{{{x}}^{{2}}-{4}{x}+{2}{x}-{8}}}$

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

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

Now we create the partial fraction template,

$\displaystyle\frac{{{x}+{5}}}{{{\left({x}-{4}\right)}{\left({x}+{2}\right)}}}=\frac{{A}}{{{x}-{4}}}+\frac{{B}}{{{x}+{2}}}$

Multiply the above equation with the denominator,

$\displaystyle\Rightarrow{\left({x}+{5}\right)}={A}{\left({x}+{2}\right)}+{B}{\left({x}-{4}\right)}$

$\displaystyle\Rightarrow{x}+{5}={A}{x}+{2}{A}+{B}{x}-{4}{B}$

$\displaystyle\Rightarrow{x}+{5}={\left({A}+{B}\right)}{x}+{2}{A}-{4}{B}$

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={1}$ ---------------(1)

$\displaystyle{2}{A}-{4}{B}={5}$ -------------(2)

Now we to solve the above linear equations to get the values of A and B,

Multiply equation 1 by 4,

$\displaystyle{4}{A}+{4}{B}={4}$ -------------(3)

Now add equation 2 and 3,

$\displaystyle{2}{A}+{4}{A}={5}+{4}$

$\displaystyle\Rightarrow{6}{A}={9}$

$\displaystyle\Rightarrow{A}=\frac{{9}}{{6}}$

$\displaystyle\Rightarrow{A}=\frac{{3}}{{2}}$

Plug in the value of A in equation 1 ,

$\displaystyle\frac{{3}}{{2}}+{B}={1}$

$\displaystyle\Rightarrow{B}={1}-\frac{{3}}{{2}}$

$\displaystyle\Rightarrow{B}=-\frac{{1}}{{2}}$

Plug in the values of A and B in the partial fraction template,

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

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

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}=\int{\left({2}{x}+\frac{{3}}{{{2}{\left({x}-{4}\right)}}}-\frac{{1}}{{{2}{\left({x}+{2}\right)}}}\right)}{\left.{d}{x}\right.}$

Apply the sum rule,

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

Take the constant out,

$\displaystyle={2}\int{x}{\left.{d}{x}\right.}+\frac{{3}}{{2}}\int\frac{{1}}{{{x}-{4}}}{\left.{d}{x}\right.}-\frac{{1}}{{2}}\int\frac{{1}}{{{x}+{2}}}{\left.{d}{x}\right.}$

Apply the power rule for the first integral and use the common integral $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$ for the second and third integral:

$\displaystyle={2}\frac{{{x}}^{{2}}}{{2}}+\frac{{3}}{{2}}{\ln}{\left|{x}-{4}\right|}-\frac{{1}}{{2}}{\ln}{\left|{x}+{2}\right|}$

Simplify and a constant C to the solution,

$\displaystyle={{x}}^{{2}}+\frac{{3}}{{2}}{\ln}{\left|{x}-{4}\right|}-\frac{{1}}{{2}}{\ln}{\left|{x}+{2}\right|}+{C}$

New Notes Blog

Wednesday, 25 February 2015

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}$ 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?

Wednesday, 25 February 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?

$\displaystyle\int\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}{\left.{d}{x}\right.}$ 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?