Integration (Mathematics) HELP!?

∫(2x - 1)/(-2x² + x + 3) dx

∫(2x - 1)/(-2x² + 3x - 2x + 3) dx

∫(2x - 1)/(-2x(x - 3/2) - 2(x - 3/2)) dx

∫(2x - 1)/(-2x - 2)(x - 3/2) dx

Now use partial fractions:

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

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

A=6/5 and B = -2/5

∫-3/5 * 1/(x + 1) - 2/5*1/(x - 3/2) dx

=-3/5*ln|x + 1| - 2/5*ln|x - 3/2| + C

Integral ( (2x - 1) / (3 + x - 2x^2) dx )

First, factor the denominator. Placing them in descending power order, we get

Integral ( (2x - 1) / ( -2x^2 + x + 3 ) dx )

Factoring (-1) from the denominator gives us

Integral ( (2x - 1) / [ (-1) (2x^2 - x - 3) ] dx )

Now, we can pull the (-1) from the denominator out of the integral itself. Since 1/(-1) is just -1, we get

(-1) Integral ( (2x - 1) / [ (2x^2 - x - 3) ] dx )

The denominator factors as

(-1) Integral ( (2x - 1) / [ (2x - 3)(x + 1) ] dx )

We cannot solve it like this. We must use partial fractions.

The partial fraction decomposition is:

(2x - 1) / [ (2x - 3)(x + 1) ] = A/(2x - 3) + B/(x + 1)

All we have to do is solve for A and B, and then solve this new form.

Multiply both sides of the equation by (2x - 3)(x + 1), to get

2x - 1 = A(x + 1) + B(2x - 3)

Remember that this is an equation true for all values of x, so we can make x anything we want to. In this case, we are going to choose values of x that helps us solve for A and B.

Let x = -1. Then our equation becomes

2(-1) - 1 = A(-1 + 1) + B(2(-1) - 3)

-2 - 1 = A(0) + B(-2 - 3)

-3 = 0 + B(-5)

-3 = B(-5)

3/5 = B

Now, let us make x = 3/2. That means that since

2x - 1 = A(x + 1) + B(2x - 3), we get

2(3/2) - 1 = A(3/2 + 1) + B(2(3/2) - 3)

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

2 = A(5/2) + B(0)

2 = A(5/2) + 0

2 = A(5/2)

2(2/5) = A

4/5 = A

So A = 4/5, B = 3/5, and our partial fraction decomposition is

(2x - 1) / [ (2x - 3)(x + 1) ] = (4/5) (1/(2x - 3)) + (3/5) (1/(x + 1))

And this is what we are going to integrate. As per our earlier question,

(-1) Integral ( (4/5) (1/(2x - 3)) + (3/5) (1/(x + 1)) dx )

Which we can split into two integrals, at the same time factoring the constant out, to get

(-1) [ (4/5) Integral ( 1/(2x - 3) dx ) + (3/5) Integral ( 1/(x + 1) dx )

Distributing the (-1), we get

(-4/5) Integral ( 1/(2x - 3) dx ) - (3/5) Integral ( 1/(x + 1) dx )

These are easy integrals. They are just the natural log. In the case of 1/(2x - 3), we must offset by the proper constant due to the chain rule, and we offset by (1/2).

(-4/5) (1/2) ln |2x - 3| - (3/5) ln |x + 1| + C

Reducing a bit,

(-2/5) ln|2x - 3| - (3/5) ln|x + 1| + C