Blog: `int (x^2 - 5x + 16)/((2x + 1)(x - 2)^2) dx` Evaluate the integral

$\displaystyle\int\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}{2}$

Let's first express the integrand as sum of proper rational expressions by applying partial fraction decomposition,

$\displaystyle\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{A}}{{{2}{x}+{1}}}+\frac{{B}}{{{x}-{2}}}+\frac{{C}}{{{\left({x}-{2}\right)}}^{{2}}}$

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

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

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

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

Now equate the coefficients of the polynomial in the numerator on the both sides,

$\displaystyle{A}+{2}{B}={1}$ --------------------------------(1)

$\displaystyle-{4}{A}-{3}{B}+{2}{C}=-{5}$ -----------------(2)

$\displaystyle{4}{A}-{2}{B}+{C}={16}$ ----------------------(3)

Now let's solve the above three equations by the method of substitution,

From equation 1 : $\displaystyle{A}={1}-{2}{B}$

Substitute the above value of A in equation 2 ,

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

$\displaystyle-{4}+{8}{B}-{3}{B}+{2}{C}=-{5}$

$\displaystyle{5}{B}+{2}{C}=-{5}+{4}$

$\displaystyle{5}{B}+{2}{C}=-{1}$ ------------------------ (4)

Now...

$\displaystyle\int\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}{2}$

Let's first express the integrand as sum of proper rational expressions by applying partial fraction decomposition,

$\displaystyle\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{A}}{{{2}{x}+{1}}}+\frac{{B}}{{{x}-{2}}}+\frac{{C}}{{{\left({x}-{2}\right)}}^{{2}}}$

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

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

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

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

Now equate the coefficients of the polynomial in the numerator on the both sides,

$\displaystyle{A}+{2}{B}={1}$ --------------------------------(1)

$\displaystyle-{4}{A}-{3}{B}+{2}{C}=-{5}$ -----------------(2)

$\displaystyle{4}{A}-{2}{B}+{C}={16}$ ----------------------(3)

Now let's solve the above three equations by the method of substitution,

From equation 1 : $\displaystyle{A}={1}-{2}{B}$

Substitute the above value of A in equation 2 ,

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

$\displaystyle-{4}+{8}{B}-{3}{B}+{2}{C}=-{5}$

$\displaystyle{5}{B}+{2}{C}=-{5}+{4}$

$\displaystyle{5}{B}+{2}{C}=-{1}$ ------------------------ (4)

Now substitute the value of A in equation 3,

$\displaystyle{4}{\left({1}-{2}{B}\right)}-{2}{B}+{C}={16}$

$\displaystyle{4}-{8}{B}-{2}{B}+{C}={16}$

$\displaystyle{4}-{10}{B}+{C}={16}$

$\displaystyle-{10}{B}+{C}={16}-{4}$

$\displaystyle-{10}{B}+{C}={12}$ -----------------------(5)

Now solve the equations 4 and 5 by the method of elimination,

Multiply equation 4 by 2,

$\displaystyle{10}{B}+{4}{C}=-{2}$ ----------------------(6)

Now add the equations 5 and 6,

$\displaystyle{5}{C}={12}-{2}={10}$

$\displaystyle{C}=\frac{{10}}{{5}}={2}$

Plug the value of C in equation 5.

$\displaystyle-{10}{B}+{2}={12}$

$\displaystyle-{10}{B}={12}-{2}={10}$

$\displaystyle{B}=\frac{{10}}{{-{{10}}}}=-{1}$

Plug the value of B in equation 1.

$\displaystyle{A}+{2}{\left(-{1}\right)}={1}$

$\displaystyle{A}-{2}={1}$

$\displaystyle{A}={1}+{2}={3}$

$\displaystyle\therefore\int\frac{{{{x}}^{{2}}-{5}{x}+{6}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}=\int{\left(\frac{{3}}{{{2}{x}+{1}}}+\frac{{-{1}}}{{{x}-{2}}}+\frac{{2}}{{{\left({x}-{2}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$

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

Now let's evaluate the above three integrals,

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

Let's apply the integral substitution: $\displaystyle{u}={2}{x}+{1}$

$\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{2}{u}}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{u}\right|}$

Substitute back u=2x+1,

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{2}{x}+{1}\right|}$

Now let's evaluate $\displaystyle\int\frac{{1}}{{{x}-{2}}}{\left.{d}{x}\right.}$

apply integral substitution: $\displaystyle{v}={x}-{2}$

$\displaystyle{d}{v}={\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{v}}{d}{v}$

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

substitute back v=x-2,

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

Now let's evaluate integral $\displaystyle\int\frac{{1}}{{{\left({x}-{2}\right)}}^{{2}}}{\left.{d}{x}\right.}$

apply the integral substitution: t=x-2

$\displaystyle{\left.{d}{t}\right.}={\left.{d}{x}\right.}$

$\displaystyle\int\frac{{1}}{{{t}}^{{2}}}{\left.{d}{t}\right.}$

$\displaystyle=\int{{t}}^{{-{{2}}}}{\left.{d}{t}\right.}$

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

$\displaystyle=-\frac{{1}}{{t}}$

Substitute back t=x-2,

$\displaystyle=-\frac{{1}}{{{x}-{2}}}$

$\displaystyle\therefore\int\frac{{{{x}}^{{2}}-{5}{x}+{16}}}{{{\left({2}{x}+{1}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}={3}{\left(\frac{{1}}{{2}}{\ln}{\left|{2}{x}+{1}\right|}\right)}-{1}{\ln}{\left|{x}-{2}\right|}+{2}{\left(-\frac{{1}}{{{x}-{2}}}\right)}$

$\displaystyle=\frac{{3}}{{2}}{\ln}{\left|{2}{x}+{1}\right|}-{\ln}{\left|{x}-{2}\right|}-\frac{{2}}{{{x}-{2}}}+{C}$ where C is a constant

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