Integration of Rational Functions

[Biosyn]

Homework Statement

∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx

Homework Equations

Partial functions

The Attempt at a Solution

∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx = ∫(Ax)/(x²+2x-8) + B/(x-1)

= ∫A(x-1)dx + ∫B(x^2+2x-8)dx

Or is it:
= ∫(AX+B)/(x^2-2x-8) dx + ∫ C/(x-1)dx

Have I done the separation of the numerator and denominators right? I'm stuck right about here..I don't know what Bx² equals to. 7? That doesn't seem to make sense, I must have done something wrong earlier.
Please help, thanks!

 

[SammyS]

Homework Statement

∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx

Homework Equations

Partial functions

The Attempt at a Solution

∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx = ∫(Ax)/(x²+2x-8) + B/(x-1)

= ∫A(x-1)dx + ∫B(x^2+2x-8)dx

Or is it:
= ∫(AX+B)/(x^2-2x-8) dx + ∫ C/(x-1)dx

Have I done the separation of the numerator and denominators right? I'm stuck right about here..I don't know what Bx² equals to. 7? That doesn't seem to make sense, I must have done something wrong earlier.
Please help, thanks!

Do a complete partial fraction decomposition.

$\displaystyle \frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}=\frac{A}{x-2}+\frac{B}{x+4}+\frac{C}{x-1}\,.$

 

[Biosyn]

Do a complete partial fraction decomposition.

$\displaystyle \frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}=\frac{A}{x-2}+\frac{B}{x+4}+\frac{C}{x-1}\,.$

Okay, so I did that.

= ∫ $\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}$

and this is what I get after multiplying everything out. Three equations. Now I just need to solve for A,B,C...

-4A+B-8C=-54

3A-3B+2C=22

A+B+C =7

 

[SammyS]

Okay, so I did that.

= ∫ $\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}$  equals What?

and this is what I get after multiplying everything out. Three equations. Now I just need to solve for A,B,C...

-4A+B-8C=-54

3A-3B+2C=22

A+B+C =7

Do you know how to do partial fraction decomposition?

 

[SammyS]

$\displaystyle \frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}=\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}$

So that   $\displaystyle 7x^2+22x-54=A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)\,.$

There's a neat trick to finishing this.

Let x=2 to find A.

Let x=-4 to find B.

Let x=1 to find C.

 

[Biosyn]

∫ $\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}$ = ∫ $\frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}$dx ?The values for X are the zeroes, could you explain to me how they are used to find A,B,C? (the trick you typed above)

 

[SammyS]

See post #5.

 

[Biosyn]

See post #5.

[STRIKE]Okay,

I multiplied everything out: Ax^2+3AX-4A+Bx^2-3Bx+2B+Cx^2+2Cx-8C

and ...I'm back to what I did in post #3 solving for A,B,C ?

Sorry, it's getting late and I'm getting a little frustrated, maybe I'm not thinking.[/STRIKE]nevermind, I got it now :)

A = 3
B = -1
C = 5

So the answer should be ∫f(x)dx = 3ln|x-2|-ln|x+4|+5ln|x-1| + C

I made a calculation error in post #3.

Thanks for your help Sammy! :)