Solving an Integration Problem Involving Partial Fractions and Trig Substitution

[Tilted]

Homework Statement

\int(x^2-2x+3)dx/(x^3-x^2-x-2)

Homework Equations

Trig substitution/ partial fractions?

The Attempt at a Solution

I used partial fractions to reduce the integral down to:

\intdx/(x-2)+\int(x-1)dx/(x^2+x+1)

The first integral is easy enough, but the second one I'm not sure where to start.

Thanks in advance!

 

[Dick]

I don't agree with all of your numbers there but the general form is right. For the quadratic part you want to complete the square in the denominator, do substitution for the squared part, then split it up and use a trig substitution.

 

[jzachey]

using partial fractions I obtained
(x^2-2x+3)
-----------dx
((x-2)(x^2+x+1)

A bx+c
-- + --------
x-2 x^2+x+1

A=3/7
B=4/7
c=-9/7
this is my attempt, i how it helps somwhat, I'm not sure if it's correct so I apologize early on.

 

[SammyS]

using partial fractions I obtained
(x^2-2x+3)
-----------dx
((x-2)(x^2+x+1)

A Bx+C
-- + --------
x-2 x^2+x+1

A                 Bx+C
--       +     --------
x-2           x^2+x+1

A=3/7
B=4/7
C=-9/7
this is my attempt, i how it helps somwhat, I'm not sure if it's correct so I apologize early on.

Those values for A, B, and C, are correct.

 

[jzachey]

A                 Bx+C
--       +     --------
x-2           x^2+x+1

Those values for A, B, and C, are correct.

Oh thank you so this means
∫\frac{3}{7(x-2)}+\frac{4x-9}{7(x+x+1)}
From here you integrate

 

[SammyS]

Oh thank you so this means
\displaystyle \int\left(\frac{3}{7(x-2)}+\frac{4x-9}{7(x^2+x+1)}\right)dx
From here you integrate

There's a typo in your post, corrected (and reformatted) above.