integration?

Question

integral=S

evaluate

S{(2x^3+3x^2-3)
/(2x^2-x-1)}dx

integrate the function two x
cubed plus three x squared
minus three,divided by
two x squared minus x minus
one?

the first answerer is on the
right track

Answer 1

Divide numerator by denominator to obtain
x + 2 + (3x-1) / (2x² - x - 1)

x + 2 can be integrated easily so consider
(3x - 1) / (2x² - x - 1) = (3x-1) / (2x +1)(x -1) =

A / (2x + 1) + B / (x - 1), using method of partial fractions

3x - 1 = A(x - 1) + B(2x + 1)
Let x = 1 gives B = 2/3
Let x = - (1/2) gives A = 5/3

Integral then becomes:-

Int(x + 2)dx + (5/3) Int 1/(2x + 1) dx + (2/3) Int(1/(x -1) dx

= x² + 2x + (5/6 ) log(2x + 1) + (2/3) log(x - 1) + C

Hope you can follow this.
It is difficult to show the long division at start of question on this site.
Please return if you are not comfortable with solution.

Answer 2

Divide the top by the bottom and if a fraction remains, you can either use u substitution (let u= the denominator) or partial fractions.

Dividing through and then factoring the denominator gives you:
x+2 + (3x-1)/[(x-1)(2x+1)]

Now integral of x+2 is just x^2/2 + 2x and the method of partial fractions will give you the integral of the fraction.

Basically, you want to find the constants a and b for which a/(x-1) + b/(2x+1) = (3x-1)/[(x-1)(2x+1)]...

Answer 3

i dont really feel like doing it out, too bad u dont have a TI 89 calculator, those got me through Calc A, B, C, advanced engineering math classes etc etc etc
i believe u do U substituion, where you let U = 2x^3 +3x^2-3 and du/dx would be the derivative of U

to keep the problem equal when using the u substituion, you'd have to multiply by 3 oustide the integral symbol....if u do it out u'll see what im talking about. good luck