Question

Guys, I need help on this one:
\int((5x^3-3x*2+2x-1)/(x^4-x^2))

Answer 1

Is it this:
\[\int\limits \frac{5x^3-3x^2+2x-1}{x^4-x^2}\]

Answer 2

When was the last time you warmed up your Partial Fractions?

Answer 3

@tkhunny I did yesterday, but still can't find the solution....

Answer 4

You'll need some denominators. Looks like x, x^2, x+1 and x-1.

Now what?

Answer 5

I'll retry that, please wait a moment ;)

Answer 6

@tkhunny : Still can't find the solution:
\[\frac Ax+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{x-1}=Ax(x+1)(x-1)+B(x+1)(x-1)+Cx^2(x-1)+Dx^2(x+1) \]
I found illogical solution for B, B=-1 and B=1
:?
Can you help please?

Answer 7

Did you continue to expand your expression? There aren't too many shortcuts, here.

Answer 8

"Did you continue to expand your expression?" I did expand the RHS above and equate them with 5x^3-3x*2+2x-1, that way I found B=-3 and B=1 too. Do you mean that?

Answer 9

You should get: \(x^{3}(A+C+D) + x^{2}(B+C-D) + x(-A) + (-B)\)

Did you?

Answer 10

I thought I messed up somewhere, I got \[(A+C+D)x^3+Bx^2+(-A-C+D)x-B\]

Answer 11

Better try that again. That would indeed lead to an irrational solution.

Be careful. Make an accounting system so that you are not confused.

Answer 12

Can you lead me step by step in expanding the fraction? I really don't know where I messed up :?

Answer 13

@tkhunny

Answer 14

This is a calculus problem. You should not be hung up on the algebra, tedious though it may be.

See how I wrote the variable in front of the coefficient? This is how I collect coefficients from annoyingly-long expressions. Develop some sort of scheme that benefits you. Keep trying until you get it!

Answer 15

Oh myy..... I just slupped \[Cx^2(x-1)=Cx^3-Cx\]
Thanks mate, here's your medal ;)

Answer 16

*slipped

Answer 17

So, what did you get?

A = ?
B = ?
C = ?
D = ?

There are lots more errors to make, and we haven't even gotten to the calculus.

Answer 18

Lol, okay, I'll try again:

Answer 19

\(A(x^{3} - x)\)

Answer 20

\(B(x^{2} - 1)\)

Answer 21

\(C(x^{3} + x^{2})\)

Answer 22

\(D(x^{3} - x^{2})\)

Answer 23

Start collecting!

Answer 24

A=-2, B=1, C=3/2, D=11/2. Is that it?

Answer 25

I don't know that you didn't just copy that out of a book. Shall we ignore that and get to the calculus?

Answer 26

Lol, okay, we get to calculus. I think I can do the A, C, and D part with substitution? how about the B part: \[\int{\frac{1}{x^2}}dx\]

Answer 27

There's no substitution.

Do this one first \(\int \dfrac{1}{x}\;dx\)

Answer 28

It's ln(x) +C right?

Answer 29

Not quite. That's why I asked. Without any clue about x, the correct result is

\(\int \dfrac{1}{x}\;dx = ln|x| + C\) The absolute values are not optional.

Answer 30

Ooh, any reason for the absolute mark?

Answer 31

Then for C and D, \[\frac 32 \ln{|x+1|}+\frac{11}{2}\ln|x-1|+C\] amI right?

Answer 32

@tkhunny

Answer 33

If you KNOW x > 0, you can skip the absolute values.
If you do NOT know x > 0, they are mandatory because the logarithm function blows up!

Answer 34

B is just the power rule or polynomial rule. Whatever you wish to call it.

Answer 35

Okay, thanks mate. You saved my day :)

Answer 36

You may also use your logarithm properties to simplify the expression.