Calculus Partial Fractions, Help Please...?

Question

integral of 10dx / (x-1)(x^2+9)

Answer 1

You want to express 10/(x-1)(x^2+9) as:

a/(x-1) + (bx+c)/(x^2 + 9)

To do this take a/(x-1) + bx/(x^2 + 9), put it over a common denominator and set the numerator to 10 to get:

a(x^2 + 9) + (bx+c)(x-1) = 10

Multiplying out you get:

(a + b)x^2 + (c-b)x + 9a-c = 10

So
(a+b) = 0 which tells you b = -a
(c-b) = 0 which tells you b = c
and 9a-c = 10 combined with the above tells you a=1

So you have:
10 / (x-1)(x^2+9) = 1/(x-1) - (x+1)/(x^2+9)

= 1/(x-1) - x/(x^2 + 9) - 1/(x^2+9)

The first term integrates to become ln|x-1|. The second term integrates to become (1/2)ln(x^2 + 9). To integrate the third term, let x = 3tany. Then dx becomes 3dy(secy)^2, and dx/(x^2+9) becomes dy/3. The integral of this is y/3 = (1/3)arctan(x/3).

Answer 2

take the 10 outside, break into partial fractions,
f(x) = 1/ (x+1)(x^2 +9)
at this point you may choose to factor the quadratic using complex factors. However assuming that you don't,
f(x) = A / (x+1) + (Bx+C)/(x^2 +9)
Now by plugging values for x and/or equating coefficients find the constants A,B and C.
Now whatever A may be the derivative of (x+1) appears in its numerator. This is integrated using the log function. the quadratic term can be integrated using the arctangent function.
Note however that the complex number method and the arctan method are indeed equivalent.
Hope this helps!

Answer 3

no need for complex numbers...its just A/(x-1)+(Bx+C)(x^2+9)
multiply the whole equation by the denominator to get rid of them..then solve for the unknowns by plugging in values of x. use x=1 to get rid of the A term. once you find A, use x=0 to get rid of the B term and plug in A at the same time. pick a 3rd value and plug in A and B...this will allow you to find C. then integrate!

Answer 4

This is easy using complex numbers.
Write x^2+9 as (x - 3i)(x+3i).

Then, break 10 / (x-1)(x - 3i)(x+3i) into partial fractions
A / (x-1) + B / (x - 3i) + C / (x+3i)

Then integrate each fraction.