Blog: `int 10/((x - 1)(x^2 + 9)) dx` Evaluate the integral

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

Solve for the variables A, B, and, C using the method of partial fractions.

$\displaystyle\frac{{10}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{9}\right)}}}=\frac{{A}}{{{x}-{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{9}}}$

$\displaystyle{10}={A}{\left({{x}}^{{2}}+{9}\right)}+{\left({B}{x}+{C}\right)}{\left({x}-{1}\right)}$

$\displaystyle{10}={A}{{x}}^{{2}}+{9}{A}+{B}{{x}}^{{2}}+{C}{x}-{B}{x}-{C}$

$\displaystyle{10}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left({C}-{B}\right)}{x}+{\left({9}{A}-{C}\right)}$

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

$\displaystyle{0}={A}+{B}$

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

$\displaystyle{0}={C}-{B}$

$\displaystyle{0}={C}+{A}$

$\displaystyle{10}={9}{A}-{C}$

$\displaystyle{0}={A}+{C}$

$\displaystyle{10}={10}{A}$

$\displaystyle{A}={1}$

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

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

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

$\displaystyle=\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}-\int\frac{{x}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}$

The first integral matches the form $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int\frac{{1}}{{{x}-{1}}}={\ln}{\left|{x}-{1}\right|}+{C}$

Integrate the second integral using u-substitution.

Let $\displaystyle{u}={{x}}^{{2}}+{9}$

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

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

$\displaystyle-\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=-\frac{{x}}{{u}}\cdot\frac{{{d}{u}}}{{{2}{x}}}=-\frac{{1}}{{2}}{\ln}{\left|{u}\right|}+{C}=-\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{9}\right|}+{C}$

The third integral matches the form $\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{a}}\right)}+{C}$

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

$\displaystyle=\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}-\int\frac{{x}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}$

$\displaystyle={\ln}{\left|{x}-{1}\right|}-\frac{{1}}{{2}}{\ln{{\left({{x}}^{{2}}+{9}\right)}}}-\frac{{1}}{{3}}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{3}}\right)}+{C}$

...

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

Solve for the variables A, B, and, C using the method of partial fractions.

$\displaystyle\frac{{10}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{9}\right)}}}=\frac{{A}}{{{x}-{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{9}}}$

$\displaystyle{10}={A}{\left({{x}}^{{2}}+{9}\right)}+{\left({B}{x}+{C}\right)}{\left({x}-{1}\right)}$

$\displaystyle{10}={A}{{x}}^{{2}}+{9}{A}+{B}{{x}}^{{2}}+{C}{x}-{B}{x}-{C}$

$\displaystyle{10}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left({C}-{B}\right)}{x}+{\left({9}{A}-{C}\right)}$

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

$\displaystyle{0}={A}+{B}$

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

$\displaystyle{0}={C}-{B}$

$\displaystyle{0}={C}+{A}$

$\displaystyle{10}={9}{A}-{C}$

$\displaystyle{0}={A}+{C}$

$\displaystyle{10}={10}{A}$

$\displaystyle{A}={1}$

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

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

$\displaystyle=\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}-\int\frac{{x}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}$

The first integral matches the form $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int\frac{{1}}{{{x}-{1}}}={\ln}{\left|{x}-{1}\right|}+{C}$

Integrate the second integral using u-substitution.

Let $\displaystyle{u}={{x}}^{{2}}+{9}$

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

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

The third integral matches the form $\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{a}}\right)}+{C}$

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

$\displaystyle=\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}-\int\frac{{x}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{{x}}^{{2}}+{9}}}{\left.{d}{x}\right.}$

$\displaystyle={\ln}{\left|{x}-{1}\right|}-\frac{{1}}{{2}}{\ln{{\left({{x}}^{{2}}+{9}\right)}}}-\frac{{1}}{{3}}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{3}}\right)}+{C}$

The final answer is:

$\displaystyle={\ln}{\left|{x}-{1}\right|}-\frac{{1}}{{2}}{\ln{{\left({{x}}^{{2}}+{9}\right)}}}-\frac{{1}}{{3}}{{\tan}}^{{-{{1}}}}{\left(\frac{{x}}{{3}}\right)}+{C}$

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$\displaystyle\int\frac{{10}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{9}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

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Evaluate the integral

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

$\displaystyle\int\frac{{10}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{9}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

What are the problems with Uganda's government?