Question

How do you integrate `int (3x+7)/(x^4-16)dx` using partial fractions?

Answer 1

`1/32[13 ln|x-2| -ln|x+2| - 6 ln(x^2 + 4) -14 arctan (x/2)] + C`

Explanation:

*Step 1:* Partial fraction decomposition
We begin be factor the denominator

`(3x+7)/(x^4-16)= (3x+7)/((x^2 +4)(x^2-4)) = (3x+7)/((x-2)(x+2)(x^2+4))`

`=A/(x-2)+B/(x-2) +(Cx+D)/(x^2 +4) " " " " " " " "color(red)((1))`

`3x+7=A(x+2)(x^2+4) +B(x-2)(x^2+4) " " " " " " " " + (Cx+D)(x^2-4)" " " " " "color(red)((2) "`

FOIL the partial fraction expression we have
`A(x^3 +2x^2 + 4x+8)`
`B(x^3 -2x^2 +4x-8)`
`Cx^3 +Dx^2-4Cx -4D`

`0x^3 : " " A+B + C = 0 " " " " " (Eq. 3)`
`0x^2: " " 2A-2B+D= 0 " " " " " (Eq. 4)`
`3x: " " "4A+4B-4C= 3" " " " " " " (Eq. 5)`
`7: " " " 8 A-8B -4D = 7" " " " " " " (Eq. 6)`

Step 2: Solve the system of equation above

Multiply 4 to `Eq.3` and add to `Eq.5`
`4*(A+B+C= 0) hArr 4A +4B + 4C = 0`

`+ (4A+4B-4C= 3)`
`color(red)(8 A + 8B = 3) " " " (Eq. 7)`

Multiply `(Eq. 4)` by 4 and add to `(Eq.6)`
`4(2A- 2B+D= 0) hArr 8A - 8B +4D = 0`

+`8A-8B+4D= 7`
`color(red)(16 A- 16B= 7) (Eq. 8)`

Multiply `(Eq.7)` by 2, add to `(Eq.8)`
`2(8a+8b = 3) hArr 16A -16B= 6`

`+16A-16B= 7`
`32A= 13 hArr A= 13/32 " " " " (8)`

Substitute into either `(Eq.7)` or `(Eq.8)` to solve for B
`13/32+8B= 3 hArr B= -1/32 " " " " " " (9)`

Substitute (8) and (9) into Eq. 3 and Eq.4 to solve for C and D
`C= -3/8 " " " " " (10)`
`D=-7/8 " " " " " (11)`

Step 3: Rewrite the partial faction using (1) and (8) (9) (10)(11)
`(3x+7)/(x^4-16) = 13/(32(x-2)) -(1/(32(x+2))) - (3x+7)/(8(x^2 +4)`

Step 4: Rewrite the integral
`int((3x+7)/(x^4 -16))dx = 13/32 int( 1/(x-2)) dx -1/32 int (1/(x+2)) dx - 3/8*int( x/(x^2 +4)) dx -7/8 int (1/(x^2 +4)) dx`

Step 5: Start integrating
`= 13/32 ln |x-2| -1/32 ln|x+2| -3/16 ln(x^2 +4) -7/8 arctan (x/2) +C`

`1/32[13 ln|x-2| -ln|x+2| - 6 ln(x^2 + 4) -14 arctan (x/2) ]+ C`

Sorry, this problem is really really long....

Please feel free to edit this, if you see any mistake.