# How do I solve this integral?

Integral of: (x^2 + 2x - 1)/(x^3 - x)
I keep getting it wrong but I'm not sure why.
asked Jan 12, 2013 in CALCULUS

Given ∫ (x^2+2x - 1 ) ( x^3-x ) dx

∫ (x^2+2x - 1 ) ( x^3-x ) dx = ∫ (  (x^2 ) * ( x^3) + (x^2) * (-x) + (2x) * (x^3 ) + (2x )*(-x) + (-1) (x^3) +(-1) (-x) ) dx

Product of two oposite signs is negtive

= ∫ ( x^5 )dx - ∫ ( x^3) dx + ∫ ( 2x^4) dx - ∫ ( 2x^2 ) dx  - ∫ (x^3) dx + ∫ ( -1) (-x) dx

Product of two same signs is positive

= ∫ ( x^5 )dx - ∫ ( x^3) dx + ∫ ( 2x^4) dx - ∫ ( 2x^2 ) dx  - ∫ (x^3) dx +∫ x dx

= ∫ ( x^5 )dx + ∫ ( 2x^4) dx  - 2 ∫ (x^3) dx - ∫ (2x^2 ) dx + ∫ xdx                  ( ∫ x^n dx = x^n+1 / n+1)

∫ (x^2+2x -1)( x^3-x ) dx = x^6 / 6 + 2x^5 / 5  - 2 x^4 / 4 -2 x^3 / 3 + x^2 / 2

The integral of .

Take constant terms.

-A = -1 => A = 1.

Take coefficient of x terms.

-B + C = 2 -> (1)

Take coefficient of x2 terms.

A + B + C = 1

Substituting A we get,

1 + B + C = 1

B + C = 0 -> (2)

Solve the equations (1) and (2).

=> C = 2/2 = 1.

Substitute C value in (1).

-B + C = 2

-B + 1 = 2

-B = 1 => B =-1.

Therefore

Substitute u and v values.

Therefore .