Question

How do you integrate `int (2x+1) /( (x-2)(x^2+4))` using partial fractions?

Answer 1

`((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c`

Explanation:

1. First step in this problem is understanding that the use of partial fractions is valid since the numerator is a lesser degree of power than the denominator.

2. Apply Partial Fractions:
   *Note that the denominator terms are Linear (left) and Quadratic (on the right), all this means is that they are terms that cannot be factorized further.
   Linear form = `A/(x-2)`
   Quadratic form = `(Bx+C)/(x^2+4)`

Add these two terms together and set it equal to the Numerator.
`2x+1=A/(x-2)+(Bx+C)/(x^2+4)`

Simplify by multiplying A by the denomenator of the other fraction, and the same for B:
`2x+1=A(x^2+4)+Bx+C(x-2)`

now distribute:
`2x+1=Ax^2+4A+Bx^2-2Bx+Cx-2C`

*Note: The biggest thing students confuse about partial fractions is this next step, in the left hand side you should think of it as `0x^2+2x+1` to understand that there are Zero `x^2`'s on the left.

`color(red)(0x^2)+color(blue)(2x)+color(green)(1)=color(red)(Ax^2)+color(green)(4A)+color(red)(Bx^2)-color(blue)(2Bx)+color(blue)(Cx)-color(green)(2C)`

Collect like terms:
`color(red)(x^2) rArr` `0=A+B`
`color(blue)x rArr` `2=-2B+C`
`color(green)"Integers"` `rArr` `1=4A-2C`

Solve by any method, here we will use substitution
Eq1: `0=A+B rArr [A=-B] hArr [B=-A]`
Eq2: `2=-2B+C rArr [2+2B=C]`
Eq3: `1=4A-2C rArr 1=2(2A-C) rArr [1/2=2A-C]`

Substitute in Eq1 and Eq2 into Eq3 to get:
`1/2 = 2(-B)-(2+2B)`
Simplify:
`1/2=-4B-2`

Solve for B:
`B=-5/8`
Use B to solve for A
`A=5/8`
Use B to solve for C
`2+2(-5/8)=C`
`C=3/4`

Plug in A, B, and C values into original fraction:
`A/(x-2)+(Bx+C)/(x^2+4)`

`(5/8)/(x-2)+(-5/8x+3/4)/(x^2+4)`

Simplify into:
`5/(8(x-2))+(-5x+6)/(8(x^2+4))`

New integral is now represented by:
`int5/(8(x-2))dx+int(-5x+6)/(8(x^2+4))dx`

Remove constants to prepare for integration:
`5*int1/(8(x-2))dx+1/8*int(-5x+6)/((x^2+4))dx`

Break apart second integral (on the right) into two separate integrals (Mind the 1/8 multiplier) and remove the constants in the numerators

`5/8*int1/((x-2))dx+1/8*(-5*intx/(x^2+4)dx+6*int1/(x^2+4)dx)`

First integral:
`5/8int1/(x-2)dx`
`u=x-2`
Result is: `(5ln|x-2|)/8`

Second Integral:
`5*intx/(x^2+4)dx`
`u=x^2+4`
Result Is: `(-5ln|x^2+4|)/2`

Third Integral:
`6*int1/(x^2+4)dx`
Apply inverse trig formula by identifying denominator as Arctan:
`int1/(x^2+a^2)dx = 1/a * tan^-1(x/a)`
Where `[a^2=4] rArr [a=2]`
Result is: `3tan^-1(x/2)`

Plug these back into full equation and do not forget the `1/8` multiplier for the second and third integrals.

`(5*ln|x-2|)/8-1/8((-5ln|x^2+4|)/2+3tan^-1(x/2))`

Distribute the `1/8` multiplier

`(5*ln|x-2|)/8+(5ln|x-2|)/16-(3tan^-1(x/2))/8+c`

Simplify:
`((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c`

You can simplify further if you like, however this is an acceptable answer.