New Notes Blog: `int (x^2+6x+4)/(x^4+8x^2 + 16) dx` Use partial fractions to find the indefinite integral

Saturday, 6 June 2015

$\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

$\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$

To solve using partial fraction method, the denominator of the integrand should be factored.

$\displaystyle\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}=\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

If the factor in the denominator is quadratic and repeating, the partial fraction of this factor is $\displaystyle\frac{{{A}_{{1}}{x}+{B}_{{1}}}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}+\frac{{{A}_{{2}}{x}+{B}_{{2}}}}{{{\left({a}{{x}}^{{2}}+{b}{x}+{c}\right)}}^{{2}}}+\ldots+\frac{{{A}_{{n}}{x}+{B}_{{n}}}}{{{\left({a}{{x}}^{{2}}+{b}{x}+{C}\right)}}^{{n}}}$ .

So expressing the integrand as sum of fractions, it becomes:

$\displaystyle\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}=\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{4}}}+\frac{{{C}{x}+{D}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$

To solve for the values of A, B, C and D, multiply both sides by the LCD.

$\displaystyle{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}\cdot\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}={\left(\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{4}}}+\frac{{{C}{x}+{D}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}\right)}\cdot{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}$

$\displaystyle{{x}}^{{2}}+{6}{x}+{4}={\left({A}{x}+{B}\right)}{\left({{x}}^{{2}}+{4}\right)}+{C}{x}+{D}$

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

At the right side, group together the terms with same power of x.

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

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

Notice that the right side has a degree of 3. So express the polynomial at the left side with a degree of 3.

$\displaystyle{0}{{x}}^{{3}}+{{x}}^{{2}}+{6}{x}+{4}={A}{{x}}^{{3}}+{B}{{x}}^{{2}}+{\left({4}{A}+{C}\right)}{x}+{\left({4}{B}+{D}\right)}$

In order that the two polynomials to be equal, the coefficients and the constant should be the same.

So set the coefficient of $\displaystyle{{x}}^{{3}}$ at the left side equal to the coefficient of $\displaystyle{{x}}^{{3}}$ at the right side.

$\displaystyle{0}={A}$

Also, set the coefficient of $\displaystyle{{x}}^{{2}}$ at the left side equal to the coefficient of $\displaystyle{{x}}^{{2}}$ at the right side.

$\displaystyle{1}={B}$

Set the coefficient of x at the left side equal to the coefficient of x at the right side too.

$\displaystyle{6}={4}{A}+{C}$ (Let this be EQ1.)

And set the constant at the left side equal to the constant at the right side.

$\displaystyle{4}={4}{B}+{D}$ (Let this be EQ2.)

Since the values of A and B are known already, plug-in them to equation 1 and 2 to get the values of C and D.

Plug-in A=0 to EQ1 to get the value of C.

$\displaystyle{6}={4}{\left({0}\right)}+{C}$

$\displaystyle{6}={C}$

And, plug-in B = 1 to EQ2 to get the value of D.

$\displaystyle{4}={4}{\left({1}\right)}+{D}$

$\displaystyle{4}={4}+{D}$

$\displaystyle{0}={D}$

So the partial fraction decomposition of the integrand is:

$\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left(\frac{{1}}{{{{x}}^{{2}}+{4}}}+\frac{{{6}{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$

Expressing it as sum of two integrals, it becomes:

$\displaystyle=\int\frac{{1}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}+\int\frac{{{6}{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}+{6}\int\frac{{{x}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}{\left.{d}{x}\right.}$

For the first integral, apply the formula $\displaystyle\int\frac{{1}}{{{{u}}^{{2}}+{{a}}^{{2}}}}{d}{u}=\frac{{1}}{{a}}{{\tan}}^{{-{1}}}{\left(\frac{{u}}{{a}}\right)}+{C}$ .

$\displaystyle{u}={x}$

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

$\displaystyle{a}={2}$

For the second integral, apply the formula $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle{u}={{x}}^{{2}}+{4}$

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

So the result of each integral is:

$\displaystyle=\int\frac{{1}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}+{6}\int{{\left({{x}}^{{2}}+{4}\right)}}^{{-{2}}}\cdot{x}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}+{3}\int{{\left({{x}}^{{2}}+{4}\right)}}^{{-{2}}}\cdot{2}{x}{\left.{d}{x}\right.}$

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

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

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

Therefore, $\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{x}}{{2}}\right)}-\frac{{3}}{{{{x}}^{{2}}+{4}}}+{C}$ .

New Notes Blog

Saturday, 6 June 2015

$\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ 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?

Saturday, 6 June 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?

$\displaystyle\int\frac{{{{x}}^{{2}}+{6}{x}+{4}}}{{{{x}}^{{4}}+{8}{{x}}^{{2}}+{16}}}{\left.{d}{x}\right.}$ 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?