Integral Help: Solving a Complex Problem

[tylersmith7690]

Integral help!

Homework Statement

Evaluate the following inde nite integral,
∫$\frac{x^3-5x}{x^3-2x^2+4x-8}$ dx

Homework Equations

I think I am right until i get to equating the equations when I do it the method I am taught in class I get c=0 and I don't think that is right. However when i put it into a system of equations with the coefficients i get fractions.

I would like to know what to do next. Thanks

The Attempt at a Solution

First the degree of numerator is same as the denomenator. So i do long division and end up with

∫1+ $\frac{2x^2-9x+8}{x^3-2x^2+4x-8}$

Now decompose the fraction part into:

$\frac{2x^2-9x+8}{(x^2+4)(x-2)}$ =$\frac{Ax+B}{(x^2+4)}$+$\frac{C}{(x-2)}$

Then get a common denomenator for RHS
= $\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}$

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Kind regards

 

[CAF123]

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Those values of A,B and C are correct. However, in your initial attempt, you should not get C=0.

 

[HallsofIvy]

Homework Statement

Evaluate the following inde nite integral,
∫$\frac{x^3-5x}{x^3-2x^2+4x-8}$ dx

Homework Equations

I think I am right until i get to equating the equations when I do it the method I am taught in class I get c=0 and I don't think that is right. However when i put it into a system of equations with the coefficients i get fractions.

I would like to know what to do next. Thanks

The Attempt at a Solution

First the degree of numerator is same as the denomenator. So i do long division and end up with

∫1+ $\frac{2x^2-9x+8}{x^3-2x^2+4x-8}$

Now decompose the fraction part into:

$\frac{2x^2-9x+8}{(x^2+4)(x-2)}$ =$\frac{Ax+B}{(x^2+4)}$+$\frac{C}{(x-2)}$

Then get a common denomenator for RHS
= $\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}$

So the two numerators are equal: $2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)$

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.

Taking x= 2 gives 8- 18+ 8= 8C which gives C= -1/4, as you have below, not 0. Did you forget the left side of the equation?

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Kind regards

If, in $2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)$, you take x= 0 (just because it is an easy number) you get $8= (0+ B)(-2)+ C(4)$ or $-2B= -8- (-1/4)(4)= -9$ so that C= -9/2, again as you say. Finally, taking x= 1 (again just because it is an easy number) gives $2- 9+ 8= (A+ B)(-1)+ C(5)$ or $1= -A+ 9/2- 5/4$ so that $A= 9/2- 1- 5/4= 18/4- 4/4- 5/4= 9/4$, again, exactly what you have.

Great! Now all you need to do is the integral:
$$\int dx+ \dfrac{9}{4}\int \dfrac{xdx}{x^2+ 4}- \dfrac{9}{2}\int \dfrac{dx}{x^2+ 4}- \dfrac{1}{4}\int \dfrac{dx}{x- 2}$$

 

[tylersmith7690]

Thanks for the reply, looking again at how i got C=0 was most likely due to it being very late at night :).
But yes I find it easier solving for A,B,C in the system of equations. I see now that my thinking was on the right track and it was just the small error that through me off. Thanks for the help.

 

[tylersmith7690]

Is my final answer correct?

x + $\frac{9}{8}$ log | x²+4 | - $\frac{9}{4}$ arctan($\frac{x}{2}$) - $\frac{1}{4}$ log | x-2 | + C