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Thought-Process to Discover Knowledge

Welcome to **nub****trek**.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

Welcome to **nub****trek**.

The content is presented in small-focused learning units to enable you to

think,

figure-out, &

learn.

**Just keep tapping** (or clicking) on the content to continue in the trail and learn. continue

The content is presented in small-focused learning units to enable you to

think,

figure-out, &

learn.

To make best use of nubtrek, understand what is available.

nubtrek is designed to explain mathematics and science for young readers. Every topic consists of four sections.

nub,

trek,

jogger,

exercise.

This captures the small-core of concept in simple-plain English. The objective is to make the learner to think about. continue

Trekking is bit hard, requiring one to sweat and exert. The benefits of taking the steps are awesome. In the trek, concepts are explained with exploratory questions and your thinking process is honed step by step. continue

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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)`

*plain and simple summary*

nub

*plain and simple summary*

nub

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*simple steps to build the foundation*

trek

*simple steps to build the foundation*

trek

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

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*

Progress

*About you*

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"