Integration?

Question

Find:

{Integration sign}[(4x-10)/(5x-x^2)]dx

Thanks for your help!

Answer 1

∫ ( (4x - 10)/(5x - x²) dx )

First, let's factor (-1) from the denominator, and 2 from the numerator

∫ ( (2)(4x - 10)/[(-1)(x² - 5x)] dx )

Now, let's pull it out of the integral.

(2)(-1) ∫ ( (2x - 5)/(x² - 5x) dx )

We're going to use substitution. To make it obvious, however, I'm going to take an intermediate step and move the 2x - 5 next to the dx.

(-2) * ∫ ( 1/(x² - 5x) (2x - 5) dx )

Let u = x² - 5x. Then
du = (2x - 5) dx

(-2) * ∫ ( 1/u du )

Which is now a trivial integral.

(-2) ln|u| + C

Back-substitute u = x² - 5x to obtain

(-2) ln|x² - 5x| + C

Answer 2

Ok, let's go on:

integrate[(4x-10)/(5x-x^2)]dx

Here we need to use, partial fractions :

integrate[4x - 10] / x*(5 - x)]dx

4x - 10 / x*(5-x) = A / x + B / 5 - x = 4x - 10 / x*(5-x)

Let's find : A and B :

5A - xA + xB = 4x - 10

x(B - A) + 5A = 4X - 10

A = -2 : B = 2

THEN :

4x - 10 / x*(5-x) = -2 / x + 2 / 5-x

replacing :

integrate( -2 / x + 2 / 5-x)dx

-2*integrate(1 / x) + 2*integrate(5-x)

-2*lnx - 2*ln(5-x)

-2*(lnx + lnx(5-x))

-2*(ln(5-x)*x))

Hope that helps

Answer 3

(4x - 10) / (x.(5 - x) = A / x + B / (5 - x)
4x - 10 = A.(5 - x) + B x
When x = 5 , 10 = 5B and B = 2
When x = 0 , -10 = 5A and A = - 2
I = - ∫ (2/x).dx + 2 ∫ (1/(5 - x) dx
I = - 2 log x - 2 log (5 - x) + log K
I = - 2 [ log x + log (5 - x) ] + log K
I = log [ x.(5 - x) ]^(- 2) + log K
I = log K / [x.(5 - x)] ²