The Notes: `int(x-15)/((x^2+2x+5)(x^2+6x+10))dx` Please, help with this integral! I've been struggling for three days. It's partial fractions...

Wednesday, January 11, 2012

$\displaystyle\int\frac{{{x}-{15}}}{{{\left({{x}}^{{2}}+{2}{x}+{5}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}}}{\left.{d}{x}\right.}$ Please, help with this integral! I've been struggling for three days. It's partial fractions...

We are asked to find $\displaystyle\int\frac{{{x}-{15}}}{{{\left({{x}}^{{2}}+{2}{x}+{5}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}}}{\left.{d}{x}\right.}$

We will use partial fractions to rewrite the integrand:

$\displaystyle\frac{{{x}-{15}}}{{{\left({{x}}^{{2}}+{2}{x}+{5}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}}}=\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{2}{x}+{5}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}+{6}{x}+{10}}}$

Now multiply both sides of the equation by the LCD to get:

$\displaystyle{x}-{15}={\left({A}{x}+{B}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}+{\left({C}{x}+{D}\right)}{\left({{x}}^{{2}}+{2}{x}+{5}\right)}$

Distributing and collecting like terms we get:

$\displaystyle{x}-{15}={\left({A}+{C}\right)}{{x}}^{{3}}+{\left({6}{A}+{B}+{2}{C}+{D}\right)}{{x}}^{{2}}+{\left({10}{A}+{6}{B}+{5}{C}+{2}{D}\right)}{x}+{\left({10}{B}+{5}{D}\right)}$

Equating the coefficients we get A+C=0,6A+B+2C+D=0

10A+6B+5C+2D=1 and 10B+5D=-15

Solving the system yields A=B=1,C=-1,D=-5. Thus

$\displaystyle\int\frac{{{x}+{1}}}{{{{x}}^{{2}}+{2}{x}+{5}}}{\left.{d}{x}\right.}-\int\frac{{{x}+{5}}}{{{{x}}^{{2}}+{6}{x}+{10}}}{\left.{d}{x}\right.}$

The first integral can be done by letting $\displaystyle{u}={{x}}^{{2}}+{2}{x}+{5}$ and $\displaystyle{d}{u}={2}{x}+{2}{\left.{d}{x}\right.}$ so

$\displaystyle\int\frac{{{x}+{1}}}{{{{x}}^{{2}}+{2}{x}+{5}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}\int\frac{{{d}{u}}}{{u}}=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}{x}+{5}\right|}$

Rewrite the second integral as:

$\displaystyle\int\frac{{{x}+{5}}}{{{{x}}^{{2}}+{6}{x}+{10}}}{\left.{d}{x}\right.}=\int\frac{{{2}{x}+{6}}}{{{{x}}^{{2}}+{6}{x}+{10}}}{\left.{d}{x}\right.}-{4}\int\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{2}}+{6}{x}+{10}}}$

Then we get:

$\displaystyle\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{6}{x}+{10}\right|}$

and

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

Thus:

$\displaystyle\int\frac{{{x}-{15}}}{{{\left({{x}}^{{2}}+{2}{x}+{5}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}}}{\left.{d}{x}\right.}='$

$\displaystyle\frac{{1}}{{2}}{\left({\ln}{\left|{{x}}^{{2}}+{2}{x}+{5}\right|}-{\ln}{\left|{{x}}^{{2}}+{6}{x}+{10}\right|}\right)}-{4}{{\tan}}^{{-{1}}}{\left({x}+{3}\right)}+{C}$

The Notes

Wednesday, January 11, 2012

$\displaystyle\int\frac{{{x}-{15}}}{{{\left({{x}}^{{2}}+{2}{x}+{5}\right)}{\left({{x}}^{{2}}+{6}{x}+{10}\right)}}}{\left.{d}{x}\right.}$ Please, help with this integral! I've been struggling for three days. It's partial fractions...

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