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mathsIntegral CalculusVarious Forms and Results of Integrals

### Integration by Partial Fraction

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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

So1/(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".

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