Evaluate the integral (3x^2-16x-19)/(x^3-4x^2-3x+18)
Have you done the method of partial fractions yet?
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.
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
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
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.
To evaluate the integral of a rational function, you need to factor the denominator as much as possible to express it in terms of simpler functions.
Let's factor the denominator, x^3 - 4x^2 - 3x + 18, using synthetic division or any other method:
x^3 - 4x^2 - 3x + 18 = (x - 3)(x^2 - x - 6)
Now the denominator is factored as (x - 3)(x + 2)(x - 3).
Next, we can express the rational function as a sum of partial fractions:
(3x^2 - 16x - 19)/(x^3 - 4x^2 - 3x + 18) = A/(x - 3) + B/(x + 2) + C/(x - 3)
To find the values of A, B, and C, we can multiply both sides of the equation by the denominator:
(3x^2 - 16x - 19) = A(x + 2)(x - 3) + B(x - 3)(x - 3) + C(x + 2)
Expanding the right side:
3x^2 - 16x - 19 = (A + B)x^2 + (-4A - 4B - C)x + (-6A + 9B + 2C)
Comparing the coefficients of each term, we get the following system of equations:
A + B = 3
-4A - 4B - C = -16
-6A + 9B + 2C = -19
Solving this system of equations will give us the values of A, B, and C.
Once we determine the values of A, B, and C, we can rewrite the integral as:
∫ (3x^2 - 16x - 19)/(x^3 - 4x^2 - 3x + 18) dx = ∫ (A/(x - 3) + B/(x + 2) + C/(x - 3)) dx
Now we can integrate each term separately:
∫ A/(x - 3) dx + ∫ B/(x + 2) dx + ∫ C/(x - 3) dx
The integral of 1/(x - a) with respect to x is ln|x - a|, so the first and third terms become:
A ln|x - 3| + C ln|x - 3|
The integral of 1/(x + a) with respect to x is ln|x + a|, so the second term becomes:
B ln|x + 2|
Therefore, the final result of the integral is:
A ln|x - 3| + B ln|x + 2| + C ln|x - 3| + constant, where A, B, and C are the values obtained by solving the system of equations.