maths > integral-calculus

Integration by Partial Fraction

what you'll learn...

Overview

Integration Methods :

»  Standard forms with simple substitution

$\frac{1}{ax±b}$$\frac{1}{a x \pm b}$ : use $\frac{1}{y}$$\frac{1}{y}$ form

$\frac{1}{{\left(ax±b\right)}^{n}}$$\frac{1}{a x \pm b} ^ n$ use ${y}^{-n}$${y}^{- n}$ form

$\frac{1}{\sqrt{ax±b}}$$\frac{1}{\sqrt{a x \pm b}}$ use ${y}^{-\frac{1}{2}}$${y}^{- \frac{1}{2}}$ form

»  Standard forms with trigonometric substitution

$\frac{1}{{x}^{2}+{a}^{2}}$$\frac{1}{{x}^{2} + {a}^{2}}$ use trigonometric identity ${\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta$${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

$\frac{1}{\sqrt{{x}^{2}+{a}^{2}}}$$\frac{1}{\sqrt{{x}^{2} + {a}^{2}}}$ use trigonometric identity ${\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta$${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

$\frac{1}{\sqrt{{x}^{2}-{a}^{2}}}$$\frac{1}{\sqrt{{x}^{2} - {a}^{2}}}$ use trigonometric identity ${\mathrm{sec}}^{2}\theta -1={\mathrm{tan}}^{2}\theta$${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$

$\frac{1}{\sqrt{{a}^{2}-{x}^{2}}}$$\frac{1}{\sqrt{{a}^{2} - {x}^{2}}}$ use trigonometric identity $1-{\mathrm{sin}}^{2}\theta ={\mathrm{cos}}^{2}\theta$$1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

»  Integration by Partial Fraction

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

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

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

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

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

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

»  Integration by Partial Fraction, when denominator cannot be factorized

$\frac{1}{\sqrt{a{x}^{2}+bx+c}}$$\frac{1}{\sqrt{a {x}^{2} + b x + c}}$ convert denominator to ${y}^{2}±{k}^{2}$${y}^{2} \pm {k}^{2}$ form

$\frac{1}{a{x}^{2}+bx+c}$$\frac{1}{a {x}^{2} + b x + c}$ convert denominator to ${y}^{2}±{k}^{2}$${y}^{2} \pm {k}^{2}$ form

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

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

substitution example

To integrate $\int \frac{1}{3x+4}dx$$\int \frac{1}{3 x + 4} \mathrm{dx}$, substitute $y=3x+4$$y = 3 x + 4$

Integrating $\int \frac{1}{3x+4}dx$$\int \frac{1}{3 x + 4} \mathrm{dx}$ by substitution.

$\int \frac{1}{3x+4}dx$$\int \frac{1}{3 x + 4} \mathrm{dx}$

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

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

substitution again

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

Integrating

$\int \frac{1}{{\left(3x+4\right)}^{3}}dx$$\int \frac{1}{{\left(3 x + 4\right)}^{3}} \mathrm{dx}$ by substitution

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

using $\int {y}^{n}dy=\left(n+1\right){y}^{n+1}+c$$\int {y}^{n} \mathrm{dy} = \left(n + 1\right) {y}^{n + 1} + c$
$=\frac{-2}{3}\frac{1}{{\left(3x+4\right)}^{2}}+c$$= \frac{- 2}{3} \frac{1}{{\left(3 x + 4\right)}^{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}+3x+2}$$\int \frac{1}{{x}^{2} + 3 x + 2}$ to solve this, "factorize the polynomial and convert the ratio into sum of ratios".

Integrating $\int \frac{1}{{x}^{2}+3x+2}dx$$\int \frac{1}{{x}^{2} + 3 x + 2} \mathrm{dx}$

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

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

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

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

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

This can be integrated.

example

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

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

summary

Standard forms of integration in division by polynomial: simple substitution

$\frac{1}{ax±b}$$\frac{1}{a x \pm b}$ : use $\frac{1}{y}$$\frac{1}{y}$ form

$\frac{1}{{\left(ax±b\right)}^{n}}$$\frac{1}{a x \pm b} ^ n$ use ${y}^{-n}$${y}^{- n}$ form

$\frac{1}{\sqrt{ax±b}}$$\frac{1}{\sqrt{a x \pm b}}$ use ${y}^{-\frac{1}{2}}$${y}^{- \frac{1}{2}}$ form

Standard forms of integration in division by polynomial: trigonometric substitution

$\frac{1}{{x}^{2}+{a}^{2}}$$\frac{1}{{x}^{2} + {a}^{2}}$ use trigonometric identity ${\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta$${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

$\frac{1}{\sqrt{{x}^{2}+{a}^{2}}}$$\frac{1}{\sqrt{{x}^{2} + {a}^{2}}}$ use trigonometric identity ${\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta$${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

$\frac{1}{\sqrt{{x}^{2}-{a}^{2}}}$$\frac{1}{\sqrt{{x}^{2} - {a}^{2}}}$ use trigonometric identity ${\mathrm{sec}}^{2}\theta -1={\mathrm{tan}}^{2}\theta$${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$

$\frac{1}{\sqrt{{a}^{2}-{x}^{2}}}$$\frac{1}{\sqrt{{a}^{2} - {x}^{2}}}$ use trigonometric identity $1-{\mathrm{sin}}^{2}\theta ={\mathrm{cos}}^{2}\theta$$1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

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

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

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

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

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

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

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

when denominator cannot be factorized.

$\frac{1}{\sqrt{a{x}^{2}+bx+c}}$$\frac{1}{\sqrt{a {x}^{2} + b x + c}}$ convert denominator to ${y}^{2}±{k}^{2}$${y}^{2} \pm {k}^{2}$ form

$\frac{1}{a{x}^{2}+bx+c}$$\frac{1}{a {x}^{2} + b x + c}$ convert denominator to ${y}^{2}±{k}^{2}$${y}^{2} \pm {k}^{2}$ form

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

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

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