Blog: `int_1^2 (4y^2 - 7y - 12)/(y(y + 2)(y - 3)) dy` Evaluate the integral

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}$

To solve this, apply partial fraction decomposition.

When converting the integrand to sum of proper rational expressions, set the equation as follows:

$\displaystyle\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}=\frac{{A}}{{y}}+\frac{{B}}{{{y}+{2}}}+\frac{{C}}{{{y}-{3}}}$

Multiply both sides by the LCD.

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={A}{\left({y}+{2}\right)}{\left({y}-{3}\right)}+{B}{y}{\left({y}-{3}\right)}+{C}{y}{\left({y}+{2}\right)}$

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={A}{{y}}^{{2}}-{A}{y}-{6}{A}+{B}{{y}}^{{2}}-{3}{B}{y}+{C}{{y}}^{{2}}+{2}{C}{y}$

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={\left({A}+{B}+{C}\right)}{{y}}^{{2}}+{\left(-{A}-{3}{B}+{2}{C}\right)}{Y}-{6}{A}$

For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.

y^2:

$\displaystyle{4}={A}+{B}+{C}$ (Let this be EQ1.)

$\displaystyle-{7}=-{A}-{3}{B}+{2}{C}$ (Let...

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}$

To solve this, apply partial fraction decomposition.

When converting the integrand to sum of proper rational expressions, set the equation as follows:

$\displaystyle\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}=\frac{{A}}{{y}}+\frac{{B}}{{{y}+{2}}}+\frac{{C}}{{{y}-{3}}}$

Multiply both sides by the LCD.

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={A}{\left({y}+{2}\right)}{\left({y}-{3}\right)}+{B}{y}{\left({y}-{3}\right)}+{C}{y}{\left({y}+{2}\right)}$

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={A}{{y}}^{{2}}-{A}{y}-{6}{A}+{B}{{y}}^{{2}}-{3}{B}{y}+{C}{{y}}^{{2}}+{2}{C}{y}$

$\displaystyle{4}{{y}}^{{2}}-{7}{y}-{12}={\left({A}+{B}+{C}\right)}{{y}}^{{2}}+{\left(-{A}-{3}{B}+{2}{C}\right)}{Y}-{6}{A}$

For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.

y^2:

$\displaystyle{4}={A}+{B}+{C}$ (Let this be EQ1.)

$\displaystyle-{7}=-{A}-{3}{B}+{2}{C}$ (Let this be EQ2.)

Constant:

$\displaystyle-{12}=-{6}{A}$ (Let this be EQ3.)

To solve for the value of A, consider EQ3.

$\displaystyle-{12}=-{6}{A}$

$\displaystyle{2}={A}$

Plug-in this value of A to EQ1.

$\displaystyle{4}={A}+{B}+{C}$

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

$\displaystyle{2}={B}+{C}$

Then, isolate the C.

$\displaystyle{2}-{B}={C}$

Plug-in this expression and the value of A to EQ2.

$\displaystyle-{7}=-{A}-{3}{B}+{2}{C}$

$\displaystyle-{7}=-{2}-{3}{B}+{2}{\left({2}-{B}\right)}$

$\displaystyle-{7}=-{2}-{3}{B}+{4}-{2}{B}$

$\displaystyle-{7}={2}-{5}{B}$

$\displaystyle-{9}=-{5}{B}$

$\displaystyle\frac{{9}}{{5}}={B}$

And plug-in the value of A and B to EQ1.

$\displaystyle{4}={A}+{B}+{C}$

$\displaystyle{4}={2}+\frac{{9}}{{5}}+{C}$

$\displaystyle{4}=\frac{{19}}{{5}}+{C}$

$\displaystyle\frac{{1}}{{5}}={C}$

So the partial fraction decomposition of the integrand is:

$\displaystyle\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}=\frac{{2}}{{y}}+\frac{{\frac{{9}}{{5}}}}{{{y}+{2}}}+\frac{{\frac{{1}}{{5}}}}{{{y}-{3}}}=\frac{{2}}{{y}}+\frac{{9}}{{{5}{\left({y}+{2}\right)}}}+\frac{{1}}{{{5}{\left({y}-{3}\right)}}}$

Taking the integral of this result to:

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}$

$\displaystyle={\int_{{1}}^{{2}}}{\left(\frac{{2}}{{y}}+\frac{{9}}{{{5}{\left({y}+{2}\right)}}}+\frac{{1}}{{{5}{\left({y}-{3}\right)}}}\right)}{\left.{d}{y}\right.}$

$\displaystyle={2}{\int_{{1}}^{{2}}}\frac{{1}}{{y}}{\left.{d}{y}\right.}+\frac{{9}}{{5}}{\int_{{1}}^{{2}}}\frac{{1}}{{{y}+{2}}}{\left.{d}{y}\right.}+\frac{{1}}{{5}}{\int_{{1}}^{{2}}}\frac{{1}}{{{y}-{3}}}{\left.{d}{y}\right.}$

$\displaystyle={\left({2}{\ln}{\left|{y}\right|}+\frac{{9}}{{5}}{\ln}{\left|{y}+{2}\right|}+\frac{{1}}{{5}}{\ln}{\left|{y}-{3}\right|}\right)}{{\mid}_{{1}}^{{2}}}$

$\displaystyle={\left({2}{\ln}{\left|{2}\right|}+\frac{{9}}{{5}}{\ln}{\left|{2}+{2}\right|}+\frac{{1}}{{5}}{\ln}{\left|{2}-{3}\right|}\right)}-{\left({2}{\ln}{\left|{1}\right|}+\frac{{9}}{{5}}{\ln}{\left|{1}+{2}\right|}+\frac{{1}}{{5}}{\ln}{\left|{1}-{3}\right|}\right)}$

$\displaystyle={\left({2}{\ln}{\left|{2}\right|}+\frac{{9}}{{5}}{\ln}{\left|{4}\right|}+\frac{{1}}{{5}}{\ln}{\left|-{1}\right|}\right)}-{\left({2}{\ln}{\left|{1}\right|}+\frac{{9}}{{5}}{\ln}{\left|{3}\right|}+\frac{{1}}{{5}}{\ln}{\left|-{2}\right|}\right)}$

$\displaystyle={\left({2}{\ln{{2}}}+\frac{{9}}{{5}}{{\ln{{2}}}}^{{2}}+\frac{{1}}{{5}}{\ln{{1}}}\right)}-{\left({2}{\ln{{1}}}+\frac{{9}}{{5}}{\ln{{3}}}+\frac{{1}}{{5}}{\ln{{2}}}\right)}$

$\displaystyle={\left({2}{\ln{{2}}}+\frac{{18}}{{5}}{\ln{{2}}}+{0}\right)}-{\left({0}+\frac{{9}}{{5}}{\ln{{3}}}+\frac{{1}}{{5}}{\ln{{2}}}\right)}$

$\displaystyle=\frac{{28}}{{5}}{\ln{{2}}}-{\left(\frac{{9}}{{5}}{\ln{{3}}}+\frac{{1}}{{5}}{\ln{{2}}}\right)}$

$\displaystyle=\frac{{28}}{{5}}{\ln{{2}}}-\frac{{9}}{{5}}{\ln{{3}}}-\frac{{1}}{{5}}{\ln{{2}}}$

$\displaystyle=\frac{{27}}{{5}}{\ln{{2}}}-\frac{{9}}{{5}}{\ln{{3}}}$

$\displaystyle=\frac{{9}}{{5}}{\left({3}{\ln{{2}}}-{\ln{{3}}}\right)}$

$\displaystyle=\frac{{9}}{{5}}{\left({{\ln{{2}}}}^{{3}}-{\ln{{3}}}\right)}$

$\displaystyle=\frac{{9}}{{5}}{\left({\ln{{8}}}-{\ln{{3}}}\right)}$

$\displaystyle=\frac{{9}}{{5}}{\ln{{\left(\frac{{8}}{{3}}\right)}}}$

Therefore, $\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}=\frac{{9}}{{5}}{\ln{{\left(\frac{{8}}{{3}}\right)}}}$ .

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$\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}$ Evaluate the integral

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Evaluate the integral

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What are the problems with Uganda&#39;s government?

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{4}{{y}}^{{2}}-{7}{y}-{12}}}{{{y}{\left({y}+{2}\right)}{\left({y}-{3}\right)}}}{\left.{d}{y}\right.}$ Evaluate the integral

What are the problems with Uganda's government?