New Notes Blog: `int e^x/((e^(2x)+1)(e^x-1)) dx` Use substitution and partial fractions to find the indefinite integral

Friday, 17 January 2014

$\displaystyle\int\frac{{{e}}^{{x}}}{{{\left({{e}}^{{{2}{x}}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}{\left.{d}{x}\right.}$ Use substitution and partial fractions to find the indefinite integral

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

Apply integral substitution: $\displaystyle{u}={{e}}^{{x}}$

$\displaystyle\Rightarrow{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$

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

Now let's create partial fraction template for the integrand,

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

Multiply the equation by the denominator,

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

$\displaystyle\Rightarrow{1}={A}{{u}}^{{2}}+{A}+{B}{{u}}^{{2}}-{B}{u}+{C}{u}-{C}$

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

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={0}$ -------------------------(1)

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

$\displaystyle{A}-{C}={1}$ -----------------------(3)

Now we have to solve the above three linear equations to get A, B and C,

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

Substitute B in equation 2,

$\displaystyle-{\left(-{A}\right)}+{C}={0}$

$\displaystyle\Rightarrow{A}+{C}={0}$ ---------------------(4)

Add equations 3 and 4,

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

...

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

Apply integral substitution: $\displaystyle{u}={{e}}^{{x}}$

$\displaystyle\Rightarrow{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$

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

Now let's create partial fraction template for the integrand,

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

Multiply the equation by the denominator,

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

$\displaystyle\Rightarrow{1}={A}{{u}}^{{2}}+{A}+{B}{{u}}^{{2}}-{B}{u}+{C}{u}-{C}$

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

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={0}$ -------------------------(1)

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

$\displaystyle{A}-{C}={1}$ -----------------------(3)

Now we have to solve the above three linear equations to get A, B and C,

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

Substitute B in equation 2,

$\displaystyle-{\left(-{A}\right)}+{C}={0}$

$\displaystyle\Rightarrow{A}+{C}={0}$ ---------------------(4)

Add equations 3 and 4,

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

$\displaystyle\Rightarrow{A}=\frac{{1}}{{2}}$

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

Plug in the value of A in equation 4,

$\displaystyle\frac{{1}}{{2}}+{C}={0}$

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

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

$\displaystyle\frac{{1}}{{{\left({{u}}^{{2}}+{1}\right)}{\left({u}-{1}\right)}}}=\frac{{\frac{{1}}{{2}}}}{{{u}-{1}}}+\frac{{{\left(-\frac{{1}}{{2}}\right)}{u}+{\left(-\frac{{1}}{{2}}\right)}}}{{{{u}}^{{2}}+{1}}}$

$\displaystyle=\frac{{1}}{{{2}{\left({u}-{1}\right)}}}-\frac{{{1}{\left({u}+{1}\right)}}}{{{2}{\left({{u}}^{{2}}+{1}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left[\frac{{1}}{{{u}-{1}}}-\frac{{{u}+{1}}}{{{{u}}^{{2}}+{1}}}\right]}$

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

Take the constant out,

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

Apply the sum rule,

$\displaystyle=\frac{{1}}{{2}}{\left[\int\frac{{1}}{{{u}-{1}}}{d}{u}-\int\frac{{{u}+{1}}}{{{{u}}^{{2}}+{1}}}{d}{u}\right]}$

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

Apply the sum rule for the second integral,

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

Now let's evaluate each of the above three integrals separately,

$\displaystyle\int\frac{{1}}{{{u}-{1}}}{d}{u}$

Apply integral substitution: $\displaystyle{v}={u}-{1}$

$\displaystyle{d}{v}={d}{u}$

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

Use the common integral: $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

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

Substitute back $\displaystyle{v}={u}-{1}$

$\displaystyle={\ln}{\left|{u}-{1}\right|}$ -------------------------------------------(2)

$\displaystyle\int\frac{{u}}{{{{u}}^{{2}}+{1}}}{d}{u}$

Apply integral substitution: $\displaystyle{v}={{u}}^{{2}}+{1}$

$\displaystyle{d}{v}={2}{u}{d}{u}$

$\displaystyle\int\frac{{1}}{{v}}\frac{{{d}{v}}}{{2}}$

Take the constant out and use standard integral: $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

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

Substitute back $\displaystyle{v}={{u}}^{{2}}+{1}$

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

$\displaystyle\int\frac{{1}}{{{{u}}^{{2}}+{1}}}{d}{u}$

Use the common integral: $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{{a}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{a}}{\arctan{{\left(\frac{{x}}{{a}}\right)}}}$

$\displaystyle={\arctan{{\left({u}\right)}}}$ ------------------------------------------(4)

Put the evaluation(2 , 3 and 4) of all the three integrals in (1) ,

$\displaystyle=\frac{{1}}{{2}}{\left[{\ln}{\left|{u}-{1}\right|}-\frac{{1}}{{2}}{\ln}{\left|{{u}}^{{2}}+{1}\right|}-{\arctan{{\left({u}\right)}}}\right]}$

Substitute back $\displaystyle{u}={{e}}^{{x}}$ and add a constant C to the solution,

$\displaystyle=\frac{{1}}{{2}}{\left[{\ln}{\left|{{e}}^{{x}}-{1}\right|}-\frac{{1}}{{2}}{\ln}{\left|{{e}}^{{{2}{x}}}+{1}\right|}-{\arctan{{\left({{e}}^{{x}}\right)}}}\right]}+{C}$

New Notes Blog

Friday, 17 January 2014

$\displaystyle\int\frac{{{e}}^{{x}}}{{{\left({{e}}^{{{2}{x}}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}{\left.{d}{x}\right.}$ Use substitution and 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?

Friday, 17 January 2014

Use substitution and 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?

$\displaystyle\int\frac{{{e}}^{{x}}}{{{\left({{e}}^{{{2}{x}}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}{\left.{d}{x}\right.}$ Use substitution and 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?