Evaluate the integral (3x^2-16x-19)/(x^3-4x^2-3x+18)

Answer 1

Have you done the method of partial fractions yet?

Answer 2

Yes, I know how to get variables. I got b=-8, c=1, and a=2, but I don't know how to get an integral as an answer. I need help integrating 2 other problems I will post as well, if you can help me.

Answer 3

1. (-5x^2+10x-12)/(x-5)(x^2+4)
2. (-8x-28)/((x-2)(x+9))

I know how to integrate, I just don't know how to get an integral, after finding a,b,c, and from that point

Answer 4

So you were able to use partial fractions to decompose it to

-8/(x-3)^2 + 2/(x-3) + 1/(x+2)
that was the hardest part, the rest is easy

isn't the integral of -8/(x-3)^2 equal to 8/(x-3) ?

as to the others, you should recognize the pattern of the derivative of a log function

recall that is y = ln (u)
then dy/dx = (du/dx) / ln (u)

so if we integrate
-8/(x-3)^2 + 1/(x+2) + 2/(x-3)

we get

8/(x-3) + ln(x+2) + 2ln(x-3) + constant

Answer 5

For your 2nd part,
did you get the partial fraction breakdown of

-2x/(x^2+4) - 3/(x-5) from (-5x^2 + 10x - 12)/((x-5)(x^2+4)) ?

then your integral would be

-ln(x^2+4) - 3ln(x-5) + a constant

let me know what you get for your last question.