In this page, Integration by Partial Fraction is explained with examples.

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How to integrate `int 1/(3x+4) dx`?

- Can be integrated by substituting `y=3x+4`
- Can be integrated by substituting `y=3x+4`
- cannot be integrated by substitution

The answer is "Can be integrated by substituting `y=3x+4`".

Integrating `int 1/(3x+4) dx` by substitution.

`int 1/(3x+4) dx`

`=int 1/3 (d(3x+4))/(3x+4)`* using `int (dy)/y = ln y + c` *

`=1/3 log(3x+4)+c`

How to integrate `int 1/((3x+4)^3) dx`

- Can be integrated by substituting `y=3x+4`
- Can be integrated by substituting `y=3x+4`
- even if substituted the order of denominator is 3 and so, it cannot be integrated

The answer is "Can be integrated by substituting `y=3x+4`".

Integrating

`int 1/((3x+4)^3) dx` by substitution

`=int 1/3 (d(3x+4))/((3x+4)^3)`*using `int y^n dy = (n+1)y^(n+1)+ c`*

`=(-2)/3 1/((3x+4)^2)+c`

Integration of functions that involve division by polynomial is another form that needs to be studied. For example, consider `int 1/(x^2+3x+2)`, What properties can be used to solve the integration?

- factorize the polynomial and convert the ratio into sum of ratios
- factorize the polynomial and convert the ratio into sum of ratios
- the expression cannot be integrated with algebraic manipulations

The answer is "factorize the polynomial and convert the ratio into sum of ratios".

Integrating `int 1/(x^2+3x+2) dx`

Factorizing denominator `x^2+3x+2 = (x+1)(x+2)`

Perform Partial fraction decomposition `1/(x^2+3x+2) = A/(x+1) + B/(x+2)`

Can `A` and `B` be calculated?

- yes, by equating numerators of LHS and RHS
- yes, by equating numerators of LHS and RHS
- no, it is not possible

The answer is "yes, by equating numerators of LHS and RHS".

Partial fraction decomposition:

`1/(x^2+3x+2) ``= A/(x+1) + B/(x+2)`

`1/(x^2+3x+2) ``= ((A+B)x+2A+B)/((x+1)(x+2))`* denominators are equal, and so, equate numerators to find `A` and `B`*

`A+B=0` and `2A+B=1` gives `A=1` and `B=-1`

So`1/(x^2+3x+2) ``= 1/(x+1) - 1/(x+2)`

This can be integrated.

Standard forms of integration in division by polynomial: simple substitution

`1/(ax+-b)` : use `1/y` form

`1/(ax+-b)^n` use `y^(-n)` form

`1/sqrt(ax+-b)` use `y^(-1/2)` form

Standard forms of integration in division by polynomial: trigonometric substitution

`1/(x^2+a^2)` use trigonometric identity `tan^2theta + 1 = sec^2theta`

`1/sqrt(x^2+a^2)` use trigonometric identity `tan^2theta+1 = sec^2theta`

`1/sqrt(x^2-a^2)` use trigonometric identity `sec^2theta-1 = tan^2theta`

`1/sqrt(a^2-x^2)` use trigonometric identity `1-sin^2theta = cos^2theta` *`1/(x^2-a^2)` is left out*

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

`1/(x^2-a^2)` convert to `A/(x+a) + B/(x-a)`

`(px+q)/((x+a)(x+b))` `=A/(x+a) ``+ B/(x+b)`

`(px+q)/((x+a)^2)` `=A/(x+a) ``+ B/((x+a)^2)`

`(px^2+qx+r)/((x+a)(x+b)(x+c))` `= A/(x+a) ``+ B/(x+b) ``+ C/(x+c)`

`(px^2+qx+r)/((x+a)^2(x+b))` `=A/(x+a) ``+ B/((x+a)^2) ``+ C/(x+b)`

`(px^2+qx+r)/((x+a)(x^2+bx+c))` `=A/(x+a) ``+ (Bx+C)/(x^2+bx+c)`

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

when denominator cannot be factorized.

`1/sqrt(ax^2+bx+c)` convert denominator to `y^2+-k^2` form

`1/(ax^2+bx+c)` convert denominator to `y^2+-k^2` form

`(px+q)/(ax^2+bx+c)` convert to two terms, first with numerator as derivative of denominator + second with a constant by denominator

`((ax^2+bx+c)′)/(ax^2+bx+c) ``+ A/(ax^2+bx+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

`((ax^2+bx+c)′)/sqrt(ax^2+bx+c) ``+ A/sqrt(ax^2+bx+c)`

*Solved Exercise Problem: *

Integrate `int (3x-2)/((x+1)^2(x+3))`dx

- `11/4 ln|(x+1)xx(x+3)| ``+ 5/(2(x+1)) +c`
- `11/4 ln|(x+1)/(x+3)| ``+ 5/(2(x+1)) +c`
- `11/4 ln|(x+1)/(x+3)| ``+ 5/(2(x+1)) +c`

The answer is "`11/4 ln|(x+1)/(x+3)| ``+ 5/(2(x+1)) +c`".

*slide-show version coming soon*