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Various Forms and Results of Integrals

Various Forms and Results of Integrals

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Integration Methods



 »  Standard forms with 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 with 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`



 »  Integration by Partial Fraction

   `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)`



 »  Integration by Partial Fraction, 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)`

Integration by Partial Fraction

plain and simple summary

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plain and simple summary

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


Keep tapping on the content to continue learning.
Starting on learning "". In this page, Integration by Partial Fraction is explained with examples.

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

  • 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`
  • 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
  • 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
  • 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.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

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)`



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

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`

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

your progress details

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Progress

How to integrate integral 1 by, 3x+4, dx
1
2
The answer is "can be integrated by substituting y = 3 x + 4"
Integrating integral 1 by, 3x+4, dx by substitution is given.
How to integrate integral 1 by, 3x+4,power 3, dx
1
2
The answer is "can be integrated by substituting y = 3 x + 4"
integrating integral 1 by, 3x+4, power 3, dx is given.
Integration of functions that involve division by polynomial is another form that needs to be studied. For example, consider integral 1 by, x squared + 3 x + 2. What properties can be used to solve the integration.
factorize;polynomial;convert;ratio;sum
factorize the polynomial and convert the ratio into sum of ratios
expression;cannot;not;be;b;integrated
the expression cannot be integrated with algebraic manipulations
The answer is "factorize the polynomial and convert the ratio into sum of ratios".
Integrating integral 1 by, x squared + 3 x + 2 dx. Factorizing the denominator and perform partial fraction decomposition. ;; Can A and B be calculated.
s;yes;by;equating
yes, by equating numerators of LHS and RHS
no;it;not
no, it is not possible
The answer is "yes, by equating numerators of LHS and RHS".
Partial fraction decomposition is given.
Reference Jogger: Standard forms of integration in division by polynomial: simple substitution.
Reference Jogger: Standard forms of integration in division by polynomial: trigonometric substitution.
Reference Jogger: Standard forms of integration in division by polynomial: Converting to partial fractions.
Reference Jogger: Standard forms of integration in division by polynomial: Converting to partial fractions, when denominator cannot be factorized.
Integrate the given expression.
1
2
The answer is "the second option"

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