indefinite integral?

Question

dx/(x^2-5x+6)

Answer 1

The denominator factors as (x-3)(x-2)
Now split
1/(x-3)(x-2) into partial fractions.
1/(x-3)(x-2) = A/(x-3) + B/(x-2)
1 = A(x-2) + B(x-3)
Since this is an identity, it is true for all values of x.
So let x= 3. Then A = 1.
Also let x = 2, then B = -1.
So we have reduced the integral to
∫ dx/(x-3) - ∫ dx/(x-2) = ln|x-3| - ln|x-2| + C
or ln |(x-3)/(x-2)| + C.

Answer 2

To solve this integral, you need to find the partial fraction decomposition of 1/(x^2 - 5x + 6). The denominator factors as (x - 3)(x - 2). So you want to find A and B such that

1/(x^2 - 5x + 6) = A/(x - 3) + B/(x - 2).

The common denominator is (x - 3)(x - 2), so multiply through and after cancellations you get the equation:

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

Expand:

1 = Ax - 2A + Bx - 3B

This means A + B = 0 and -2A - 3B = 1.

Solve this system of equations to get
A = 1
B = -1

Now you have

1/(x^2 - 5x + 6) = 1/(x - 3) - 1/(x - 2).

so (writing "int" for integral):

int1/(x^2 - 5x + 6)dx = int 1/(x - 3)dx - int1/(x - 2)dx

The right hand side is easy to integrate (use simple u-substitution) and the solution is ln|x - 3| - ln|x - 2| + C.

Answer 3

Use partial fractions. 1/(x^2 - 5x + 6) = 1/(x - 2)(x - 3). Let's try to write 1/(x - 2)(x - 3) as something in the form A/(x - 2) + B/(x -3).

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

Multiply both sides by (x - 2)(x - 3):

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

Now, this equation has to hold for all values of x, so you can plug in any convenient x to try and solve for A and B. Let's first plug in x = 2:

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

Now plug in x = 3:

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

So we can write 1/(x-2)(x - 3) = -1/(x - 2) + 1/(x - 3).

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

You can now integrate this directly, and get

-ln |x - 2| + ln |x - 3| + C.

Answer 4

are you asking if it is indefinite...or are you asking how to integrate it?
indefinite just means that there are no numbers listed at the top and bottom of the integral sign, resulting in having to add +C at the end of the integrated equation.