Posted by **Anonymous** on Monday, October 3, 2011 at 8:25pm.

what is the integral of (4x^2-2)/(x^2-6x-40)

- calculus -
**Reiny**, Monday, October 3, 2011 at 9:01pm
First do a long division

(4x^2-2)/(x^2-6x-40) = 4 + (24x+158)/(x^2 - 6x - 40)

now separate (24x+158)/(x^2 - 6x - 40

into partial fractions so that

A/(x-10) + B(x+4) = (24x + 158)/((x-10)(x+4))

then

A(x+4) + B(x-10) = 24x + 158

let x = -4

-14B = 62

B = -31/7

let x = 10

14A = 398

A = 199/7

(4x^2-2)/(x^2-6x-40) = 4 + (199/7)/(x-10) - (31/7)/(x+4)

so the integral is

4x + (199/7)ln(x-10) - (31/7)ln(x+10) + C , where C is a constant

- correction - calculus -
**Reiny**, Monday, October 3, 2011 at 9:12pm
I have a typo, the last line should say:

4x + (199/7)ln(x-10) - **(31/7)ln(x+4)** + C , where C is a constant

## Answer this Question

## Related Questions

- Math/Calculus - How would I solve the following integral with the substitution ...
- Calculus - Find the volume of the solid whose base is the region in the xy-plane...
- Calculus II - Integrate using integration by parts (integral) (5-x) e^3x u = 5-x...
- Calculus II/III - A. Find the integral of the following function. Integral of (x...
- calculus (please with steps and explanations) - consider the function f that is ...
- Calculus - integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is ...
- Calculus - Use the symmetry of the graphs of the sine and cosine functions as an...
- Calculus (urgent help) - consider the function f that is continuous on the ...
- calculus - 8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the...
- Calculus 2 - The question is: Evaluate the improper integral for a>0. The ...

More Related Questions