integration problem?

Question

integral of (3x^2-2) / (x^2-2x-8) dx

Answer 1

∫(3x²-2) dx / (x²-2x-8)
and we want to get only lower order terms in the numerator
= ∫(3x²-6x -24+ 6x+22) dx / (x²-2x-8)
= ∫3(x²-2x-8) dx / (x²-2x-8) + ∫(6x + 22) dx / ((x-4)(x+2))
and we now use partial fractions to get
= 3x + ∫Adx/(x-4) + ∫Bdx/(x+2)
where A = 23/3, B = -5/3
= 3x + (23/3)ln|x-4| - (5/3)ln|x+2| + C

Answer 2

Ouch. If it's any consolation, this is about as tedious an integral as you'll get in entry-level calculus.

At a minimum it looks like you're going to have to start with some partial fraction expansion. So let's put the denominator into a more useful form:

1/(x^2-2x-8) = 1/[(x-4)(x+2)] = A/(x-4) + B/(x+2)
Multiply each side by (x-4)(x+2):

1 = A(x+2) + B(x-4)

Separate common terms:

2A-4B=1;
Ax+Bx=0; A+B=0

Rewrite:

2A-4B=1;
-2A-2B=0;
-6B=1; B= -1/6; A=1/6

and 1/(x^2-2x-8) = 1/[6(x-4)] - 1/[6(x+2)]

Okay, next I will integrate each of the expanded terms separately. Let's start with term 1:

int{(3x^2-2)/6(x-4)}dx

It would be nice if this divided cleanly, but it doesn't. So I'll use a substitution instead:

u=x-4; x=u+4; du=dx

int{(3(u+4)^2-2)/6u}du
=int{(3u^2+24u+46)/6u}du
=int{u/2+4+23/3u} du
= (1/4)u^2+4u+23/3 ln(u) + C1 (C1 constant)
= (1/4)(x-4)^2+4(x-4)+23/3 ln(x-4)+C1

Okay, now for the second term:
-int{(3x^2-2)/6(x+2)}dx (I pulled the "-" in "-6" out of the denominator)

Substituting u=x+2; x=u-2; du=dx
-int{(3(u-2)^2-2)/6u}du
= -int{(3u^2-12u+10)/6u}du
= -int{(1/2)u-2+5/3u}du
= -(1/4)u^2+2u-5/3 ln(u) +C2 (C2 constant)
= -(1/4)(x+2)^2+2(x+2)-5/3 ln(x+2) +C2

So adding the terms I get

= (1/4)(x-4)^2+4(x-4)+23/3 ln(x-4)+C1 +...
-(1/4)(x+2)^2+2(x+2)-5/3 ln(x+2) +C2

This looks like it can be simplified further. In for a penny, in for a pound I guess:

=(1/4)x^2-2x+4+4x-16+23/3 ln(x-4) +...
-(1/4)x^2-x-1+2x+4-5/3 ln(x+2) + C (combining constants)

Looks like the quadratics drop out, the linears add to 3x, the zeroth powers are simply absorbed in the additive constant, and the log terms can't be simplified any further. So the final answer I get is:

3x+23/3 ln(x-4) -5/3 ln(x+2) + C

Okay, let's check it:

d/dx = 3 + 23/[3(x-4)] - 5/[3(x+2)]
The least common denominator is 3(x-4)(x+2), meaning
d/dx = 3(3(x-4)(x+2)+23(x+2)-5(x-4)) / LCD
= (9x^2-18x-72+23x+46-5x+20) / LCD
= 9x^2+(23x-18x-5x)+(46-72+20) / [3(x-4)(x+2)]
= (9x^2+0x-6)/[3(x-4)(x+2)]
= 3(3x^2-2)/[3(x-4)(x+2)]
= (3x^2-2)/(x^2-2x-8)

Woo woo!

Good luck, work hard, and stay away from drugs.

Answer 3

Divide top by bottom to obtain:-
3+ (6x+ 22) / (x² - 2x - 8)
= 3 + (6x + 22) / (x - 4).(x + 2)
(6x + 22) / (x - 4).(x + 2) = A /(x - 4) + B /(x + 2)
6x + 22 = A(x + 2) + B(x - 4)
46 = 6A
A = 46/6 = 23/3
10 = - 6B
B = (- 5/3).
I = ∫3.dx + (23/2)∫1/(x - 4).dx - (5/3).∫1/(x + 2).dx
I = 3x + (23/2).log(x - 4) - (5/3).log(x + 2) + C

Answer 4

You notice first that this fraction can be written as

3 + a/x+2 + b/x-4. You compute a and b and you are almost done. OK?