New Notes Blog: `int (x^2-1)/(x^3+x) dx` Use partial fractions to find the indefinite integral

Sunday, 27 September 2015

$\displaystyle\int\frac{{{{x}}^{{2}}-{1}}}{{{{x}}^{{3}}+{x}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

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

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

Now let's create partial fraction template,

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

Multiply equation by the denominator,

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

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

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

Comparing the coefficients of the like terms,

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

$\displaystyle{C}={0}$

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

Plug the value of A in equation 1,

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

$\displaystyle{B}={2}$

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

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

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

Apply the sum rule,

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

Take the constant out,

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

Now evaluate both the integrals separately,

$\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

Now let's evaluate second integral,

...

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

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

Now let's create partial fraction template,

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

Multiply equation by the denominator,

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

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

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

Comparing the coefficients of the like terms,

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

$\displaystyle{C}={0}$

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

Plug the value of A in equation 1,

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

$\displaystyle{B}={2}$

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

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

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

Apply the sum rule,

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

Take the constant out,

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

Now evaluate both the integrals separately,

$\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

Now let's evaluate second integral,

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

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

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

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

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

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

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

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

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

Simplify and add a constant C to the solution,

$\displaystyle=-{\ln}{\left|{x}\right|}+{\ln}{\left|{{x}}^{{2}}+{1}\right|}+{C}$

New Notes Blog

Sunday, 27 September 2015

$\displaystyle\int\frac{{{{x}}^{{2}}-{1}}}{{{{x}}^{{3}}+{x}}}{\left.{d}{x}\right.}$ Use 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?

Sunday, 27 September 2015

Use 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?

$\displaystyle\int\frac{{{{x}}^{{2}}-{1}}}{{{{x}}^{{3}}+{x}}}{\left.{d}{x}\right.}$ Use 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?