Integral Calculus : Integration by Partial Fraction

maths > integral-calculus

Integration by Partial Fraction

what you'll learn...

Overview

Integration Methods :

» Standard forms with simple substitution

   $\frac{1}{a x \pm b}$ $1/(ax+-b)$ : use $\frac{1}{y}$ $1/y$ form

   $\frac{1}{{(a x \pm b)}^{n}}$ $1/(ax+-b)^n$ use $y^{- n}$ $y^(-n)$ form

   $\frac{1}{\sqrt{a x \pm b}}$ $1/sqrt(ax+-b)$ use $y^{- \frac{1}{2}}$ $y^(-1/2)$ form

» Standard forms with trigonometric substitution

   $\frac{1}{x^{2} + a^{2}}$ $1/(x^2+a^2)$ use trigonometric identity $\tan^{2} θ + 1 = \sec^{2} θ$ $tan^2theta + 1 = sec^2theta$

   $\frac{1}{\sqrt{x^{2} + a^{2}}}$ $1/sqrt(x^2+a^2)$ use trigonometric identity $\tan^{2} θ + 1 = \sec^{2} θ$ $tan^2theta+1 = sec^2theta$

   $\frac{1}{\sqrt{x^{2} - a^{2}}}$ $1/sqrt(x^2-a^2)$ use trigonometric identity $\sec^{2} θ - 1 = \tan^{2} θ$ $sec^2theta-1 = tan^2theta$

   $\frac{1}{\sqrt{a^{2} - x^{2}}}$ $1/sqrt(a^2-x^2)$ use trigonometric identity $1 - \sin^{2} θ = \cos^{2} θ$ $1-sin^2theta = cos^2theta$

» Integration by Partial Fraction

   $\frac{1}{x^{2} - a^{2}}$ $1/(x^2-a^2)$ convert to $\frac{A}{x + a} + \frac{B}{x - a}$ $A/(x+a) + B/(x-a)$

   $\frac{p x + q}{(x + a) (x + b)}$ $(px+q)/((x+a)(x+b))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{x + b}$ $+ B/(x+b)$

   $\frac{p x + q}{{(x + a)}^{2}}$ $(px+q)/((x+a)^2)$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{{(x + a)}^{2}}$ $+ B/((x+a)^2)$

   $\frac{p x^{2} + q x + r}{(x + a) (x + b) (x + c)}$ $(px^2+qx+r)/((x+a)(x+b)(x+c))$ $= \frac{A}{x + a}$ $= A/(x+a)$ $+ \frac{B}{x + b}$ $+ B/(x+b)$ $+ \frac{C}{x + c}$ $+ C/(x+c)$

   $\frac{p x^{2} + q x + r}{{(x + a)}^{2} (x + b)}$ $(px^2+qx+r)/((x+a)^2(x+b))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{{(x + a)}^{2}}$ $+ B/((x+a)^2)$ $+ \frac{C}{x + b}$ $+ C/(x+b)$

   $\frac{p x^{2} + q x + r}{(x + a) (x^{2} + b x + c)}$ $(px^2+qx+r)/((x+a)(x^2+bx+c))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B x + C}{x^{2} + b x + c}$ $+ (Bx+C)/(x^2+bx+c)$

» Integration by Partial Fraction, when denominator cannot be factorized

   $\frac{1}{\sqrt{a x^{2} + b x + c}}$ $1/sqrt(ax^2+bx+c)$ convert denominator to $y^{2} \pm k^{2}$ $y^2+-k^2$ form

   $\frac{1}{a x^{2} + b x + c}$ $1/(ax^2+bx+c)$ convert denominator to $y^{2} \pm k^{2}$ $y^2+-k^2$ form

   $\frac{p x + q}{a x^{2} + b x + c}$ $(px+q)/(ax^2+bx+c)$ convert to two terms, first with numerator as derivative of denominator + second with a constant by denominator
$\frac{(a x^{2} + b x + c)'}{a x^{2} + b x + c}$ $((ax^2+bx+c)′)/(ax^2+bx+c)$ $+ \frac{A}{a x^{2} + b x + c}$ $+ A/(ax^2+bx+c)$

   $\frac{p x + q}{\sqrt{a x^{2} + b x + c}}$ $(px+q)/sqrt(ax^2+bx+c)$ convert to two terms, first with numerator as derivative of the quadratic equation + second with a constant by denominator
$\frac{(a x^{2} + b x + c)'}{\sqrt{a x^{2} + b x + c}}$ $((ax^2+bx+c)′)/sqrt(ax^2+bx+c)$ $+ \frac{A}{\sqrt{a x^{2} + b x + c}}$ $+ A/sqrt(ax^2+bx+c)$

substitution example

To integrate $\int \frac{1}{3 x + 4} d x$ $int 1/(3x+4) dx$ , substitute $y = 3 x + 4$ $y=3x+4$

Integrating $\int \frac{1}{3 x + 4} d x$ $int 1/(3x+4) dx$ by substitution.

$\int \frac{1}{3 x + 4} d x$ $int 1/(3x+4) dx$

$= \int \frac{1}{3} \frac{d (3 x + 4)}{3 x + 4}$ $=int 1/3 (d(3x+4))/(3x+4)$

using $\int \frac{d y}{y} = \ln y + c$ $int (dy)/y = ln y + c$
$= \frac{1}{3} \log (3 x + 4) + c$ $=1/3 log(3x+4)+c$

substitution again

To integrate $\int \frac{1}{{(3 x + 4)}^{3}} d x$ $int 1/((3x+4)^3) dx$ , substitute $y = 3 x + 4$ $y=3x+4$

Integrating

$\int \frac{1}{{(3 x + 4)}^{3}} d x$ $int 1/((3x+4)^3) dx$ by substitution

$= \int \frac{1}{3} \frac{d (3 x + 4)}{{(3 x + 4)}^{3}}$ $=int 1/3 (d(3x+4))/((3x+4)^3)$

using $\int y^{n} d y = (n + 1) y^{n + 1} + c$ $int y^n dy = (n+1)y^(n+1)+ c$
$= \frac{- 2}{3} \frac{1}{{(3 x + 4)}^{2}} + c$ $=(-2)/3 1/((3x+4)^2)+c$

extending that

Integration of functions that involve division by polynomial is another form that needs to be studied. For example, consider $\int \frac{1}{x^{2} + 3 x + 2}$ $int 1/(x^2+3x+2)$ to solve this, "factorize the polynomial and convert the ratio into sum of ratios".

Integrating $\int \frac{1}{x^{2} + 3 x + 2} d x$ $int 1/(x^2+3x+2) dx$

Factorizing denominator $x^{2} + 3 x + 2 = (x + 1) (x + 2)$ $x^2+3x+2 = (x+1)(x+2)$

Perform Partial fraction decomposition $\frac{1}{x^{2} + 3 x + 2} = \frac{A}{x + 1} + \frac{B}{x + 2}$ $1/(x^2+3x+2) = A/(x+1) + B/(x+2)$

$A$ $A$ and $B$ $B$ can be calculated by equating numerators of LHS and RHS.

naming it

Partial fraction decomposition:

$\frac{1}{x^{2} + 3 x + 2}$ $1/(x^2+3x+2)$ $= \frac{A}{x + 1} + \frac{B}{x + 2}$ $= A/(x+1) + B/(x+2)$

$\frac{1}{x^{2} + 3 x + 2}$ $1/(x^2+3x+2)$ $= \frac{(A + B) x + 2 A + B}{(x + 1) (x + 2)}$ $= ((A+B)x+2A+B)/((x+1)(x+2))$

denominators are equal, and so, equate numerators to find $A$ $A$ and $B$ $B$
$A + B = 0$ $A+B=0$ and $2 A + B = 1$ $2A+B=1$ gives $A = 1$ $A=1$ and $B = - 1$ $B=-1$

So $\frac{1}{x^{2} + 3 x + 2}$ $1/(x^2+3x+2)$ $= \frac{1}{x + 1} - \frac{1}{x + 2}$ $= 1/(x+1) - 1/(x+2)$

This can be integrated.

example

Integrate $\int \frac{3 x - 2}{{(x + 1)}^{2} (x + 3)}$ $int (3x-2)/((x+1)^2(x+3))$ dx

The answer is " $\frac{11}{4} \ln | \frac{x + 1}{x + 3} |$ $11/4 ln|(x+1)/(x+3)|$ $+ \frac{5}{2 (x + 1)} + c$ $+ 5/(2(x+1)) +c$ ".

summary

Standard forms of integration in division by polynomial: simple substitution

   $\frac{1}{a x \pm b}$ $1/(ax+-b)$ : use $\frac{1}{y}$ $1/y$ form

   $\frac{1}{{(a x \pm b)}^{n}}$ $1/(ax+-b)^n$ use $y^{- n}$ $y^(-n)$ form

   $\frac{1}{\sqrt{a x \pm b}}$ $1/sqrt(ax+-b)$ use $y^{- \frac{1}{2}}$ $y^(-1/2)$ form

Standard forms of integration in division by polynomial: trigonometric substitution

   $\frac{1}{x^{2} + a^{2}}$ $1/(x^2+a^2)$ use trigonometric identity $\tan^{2} θ + 1 = \sec^{2} θ$ $tan^2theta + 1 = sec^2theta$

   $\frac{1}{\sqrt{x^{2} + a^{2}}}$ $1/sqrt(x^2+a^2)$ use trigonometric identity $\tan^{2} θ + 1 = \sec^{2} θ$ $tan^2theta+1 = sec^2theta$

   $\frac{1}{\sqrt{x^{2} - a^{2}}}$ $1/sqrt(x^2-a^2)$ use trigonometric identity $\sec^{2} θ - 1 = \tan^{2} θ$ $sec^2theta-1 = tan^2theta$

   $\frac{1}{\sqrt{a^{2} - x^{2}}}$ $1/sqrt(a^2-x^2)$ use trigonometric identity $1 - \sin^{2} θ = \cos^{2} θ$ $1-sin^2theta = cos^2theta$

$\frac{1}{x^{2} - a^{2}}$ $1/(x^2-a^2)$ is left out

Standard forms of integration in division by polynomial: Converting to partial fractions

   $\frac{1}{x^{2} - a^{2}}$ $1/(x^2-a^2)$ convert to $\frac{A}{x + a} + \frac{B}{x - a}$ $A/(x+a) + B/(x-a)$

   $\frac{p x + q}{(x + a) (x + b)}$ $(px+q)/((x+a)(x+b))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{x + b}$ $+ B/(x+b)$

   $\frac{p x + q}{{(x + a)}^{2}}$ $(px+q)/((x+a)^2)$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{{(x + a)}^{2}}$ $+ B/((x+a)^2)$

   $\frac{p x^{2} + q x + r}{(x + a) (x + b) (x + c)}$ $(px^2+qx+r)/((x+a)(x+b)(x+c))$ $= \frac{A}{x + a}$ $= A/(x+a)$ $+ \frac{B}{x + b}$ $+ B/(x+b)$ $+ \frac{C}{x + c}$ $+ C/(x+c)$

   $\frac{p x^{2} + q x + r}{{(x + a)}^{2} (x + b)}$ $(px^2+qx+r)/((x+a)^2(x+b))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B}{{(x + a)}^{2}}$ $+ B/((x+a)^2)$ $+ \frac{C}{x + b}$ $+ C/(x+b)$

   $\frac{p x^{2} + q x + r}{(x + a) (x^{2} + b x + c)}$ $(px^2+qx+r)/((x+a)(x^2+bx+c))$ $= \frac{A}{x + a}$ $=A/(x+a)$ $+ \frac{B x + C}{x^{2} + b x + c}$ $+ (Bx+C)/(x^2+bx+c)$

Standard forms of integration in division by polynomial: Converting to partial fractions

when denominator cannot be factorized.

   $\frac{1}{\sqrt{a x^{2} + b x + c}}$ $1/sqrt(ax^2+bx+c)$ convert denominator to $y^{2} \pm k^{2}$ $y^2+-k^2$ form

   $\frac{1}{a x^{2} + b x + c}$ $1/(ax^2+bx+c)$ convert denominator to $y^{2} \pm k^{2}$ $y^2+-k^2$ form

   $\frac{p x + q}{a x^{2} + b x + c}$ $(px+q)/(ax^2+bx+c)$ convert to two terms, first with numerator as derivative of denominator + second with a constant by denominator
$\frac{(a x^{2} + b x + c)'}{a x^{2} + b x + c}$ $((ax^2+bx+c)′)/(ax^2+bx+c)$ $+ \frac{A}{a x^{2} + b x + c}$ $+ A/(ax^2+bx+c)$

   $\frac{p x + q}{\sqrt{a x^{2} + b x + c}}$ $(px+q)/sqrt(ax^2+bx+c)$ convert to two terms, first with numerator as derivative of the quadratic equation + second with a constant by denominator
$\frac{(a x^{2} + b x + c)'}{\sqrt{a x^{2} + b x + c}}$ $((ax^2+bx+c)′)/sqrt(ax^2+bx+c)$ $+ \frac{A}{\sqrt{a x^{2} + b x + c}}$ $+ A/sqrt(ax^2+bx+c)$

Outline

The outline of material to learn "Integral Calculus" is as follows.

• Detailed outline of Integral Calculus

→ Application Scenario

→ Integration First Principles

→ Graphical Meaning of Integration

→ Definition of Integrals

→ Fundamental Theorem of Calculus

→ Algebra of Integrals

→ Antiderivatives: Standard results

→ Integration of Expressions

→ Integration by Substitution

→ Integration using Identities

→ Integration by Parts

→ Integration by Partial Fraction

→ Integration: Combination of Methods