Blog: `int (x^2 + 1)/((x - 3)(x - 2)^2) dx` Evaluate the integral

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

Rewrite the rational function using partial fractions.

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

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

$\displaystyle{{x}}^{{2}}+{1}={A}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{B}{\left({{x}}^{{2}}-{5}{x}+{6}\right)}+{C}{x}-{3}{C}$

$\displaystyle{{x}}^{{2}}+{1}={A}{{x}}^{{2}}-{4}{A}{x}+{4}{A}+{B}{{x}}^{{2}}-{5}{B}{x}+{6}{B}+{C}{x}-{3}{C}$

$\displaystyle{{x}}^{{2}}+{1}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left(-{4}{A}-{5}{B}+{C}\right)}{x}+{\left({4}{A}+{6}{B}-{3}{C}\right)}$

Equate coefficients and solve for A, B, and C.

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

$\displaystyle{0}=-{4}{A}-{5}{B}+{C}$ (2)

$\displaystyle{1}={4}{A}+{6}{B}-{3}{C}$ (3)

Adding Equations (2) and (3) will give you

$\displaystyle{1}={B}-{2}{C}$

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

Using equation (1) substitute variable A with $\displaystyle{A}={1}-{B}$

into equation (3).

Using equation (4) substitute variable B with $\displaystyle{B}={1}+{2}{C}$

equation (3).

...

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

Rewrite the rational function using partial fractions.

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

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

$\displaystyle{{x}}^{{2}}+{1}={A}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{B}{\left({{x}}^{{2}}-{5}{x}+{6}\right)}+{C}{x}-{3}{C}$

$\displaystyle{{x}}^{{2}}+{1}={A}{{x}}^{{2}}-{4}{A}{x}+{4}{A}+{B}{{x}}^{{2}}-{5}{B}{x}+{6}{B}+{C}{x}-{3}{C}$

$\displaystyle{{x}}^{{2}}+{1}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left(-{4}{A}-{5}{B}+{C}\right)}{x}+{\left({4}{A}+{6}{B}-{3}{C}\right)}$

Equate coefficients and solve for A, B, and C.

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

$\displaystyle{0}=-{4}{A}-{5}{B}+{C}$ (2)

$\displaystyle{1}={4}{A}+{6}{B}-{3}{C}$ (3)

Adding Equations (2) and (3) will give you

$\displaystyle{1}={B}-{2}{C}$

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

Using equation (1) substitute variable A with $\displaystyle{A}={1}-{B}$

into equation (3).

Using equation (4) substitute variable B with $\displaystyle{B}={1}+{2}{C}$

equation (3).

$\displaystyle{1}={4}{A}+{6}{B}-{3}{C}$ (3)

$\displaystyle{1}={4}{\left({1}-{B}\right)}+{6}{\left({1}+{2}{C}\right)}-{3}{C}$

$\displaystyle{1}={4}-{4}{B}+{6}+{12}{C}-{3}{C}$

$\displaystyle{1}={10}-{4}{B}+{9}{C}$

$\displaystyle-{9}=-{4}{\left({1}+{2}{C}\right)}+{9}{C}$

$\displaystyle-{9}=-{4}-{8}{C}+{9}{C}$

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

$\displaystyle{B}={1}+{2}{C}$

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

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

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

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

$\displaystyle{A}={1}-{\left(-{9}\right)}$

$\displaystyle{A}={10}$

$\displaystyle\int\frac{{{{x}}^{{2}}+{1}}}{{{\left({x}-{3}\right)}{{\left({x}-{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}=\int\frac{{10}}{{{x}-{3}}}{\left.{d}{x}\right.}+\int-\frac{{9}}{{{x}-{2}}}{\left.{d}{x}\right.}+\int-\frac{{5}}{{{\left({x}-{2}\right)}}^{{2}}}{\left.{d}{x}\right.}$

$\displaystyle={10}{\ln}{\left|{x}-{3}\right|}-{9}{\ln}{\left|{x}-{2}\right|}+\frac{{5}}{{{x}-{2}}}+{C}$

The final answer is:

$\displaystyle={10}{\ln}{\left|{x}-{3}\right|}-{9}{\ln}{\left|{x}-{2}\right|}+\frac{{5}}{{{x}-{2}}}+{C}$

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

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Evaluate the integral

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

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

What are the problems with Uganda's government?