Indefinite integral

Question

Evaluate the following indefinite integral:

∫1/(x2-3x-10) dx

Answer 1

First you'll notice that x^2 - 3x - 10 can be factored to become (x-5)(x+2), so you're left with:

 

∫1/((x-5)(x+2)) dx

 

To evaluate the integral you'll need to break up the denominator using partial fraction decomposition. The equation you'll need to set is:

 

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

 

Multiply A by (x+2)/(x+2) and B by (x-5)/(x-5) to yield:

 

A*(x+2)/((x+2)(x-5)) + B*(x-5)/((x-5)(x+2))

 

Your equation is now:

 

1/((x-5)(x+2)) = A*(x+2)/((x+2)(x-5)) + B*(x-5)/((x-5)(x+2))

 

The left side of the equation and the right side of the equation have the same denominator, so they can be canceled. You're left with:

 

A*(x+2) + B*(x-5) = 1

 

Then you need to set your two equations to solve for A and B. First distribute A and B, then remember that 1 is the same thing as 0*x + 1

 

Ax + 2A + Bx - 5B = 0x + 1

 

Equate the x's and the constants.

 

Ax + Bx = 0x

2A - 5B = 1

 

So

 

A + B = 0

2A - 5B = 1

 

From the first equation, substitute following into the second equation:

A = -B

 

-2B - 5B = 1

-7B = 1

B = -1/7

 

Therefore,

A = 1/7

 

So now your fraction looks like this:

 

(1/7)/(x-5) + (-1/7)/(x+2)

 

Plug this back into your integral and evaluate.

 

∫(1/7)/(x-5) + (-1/7)/(x+2) dx

 

(1/7)*ln(x-5) - (1/7)*ln(x+2) + C