• Best of luck to the class of 2025 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here

Integration help (1 Viewer)

pikachu975

Premium Member
Joined
May 31, 2015
Messages
2,712
Location
NSW
Gender
Male
HSC
2017
Integrate (x-1)(x+1)dx / (x-2)(x-3)

Thanks for any help
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,078
Gender
Male
HSC
N/A
Integrate (x-1)(x+1)dx / (x-2)(x-3)

Thanks for any help
It's a quadratic over a quadratic, so use long division first and then the "remainder part" may be integrated using partial fraction decomposition (the quotient will just be a polynomial, which is easy to integrate).
 

pikachu975

Premium Member
Joined
May 31, 2015
Messages
2,712
Location
NSW
Gender
Male
HSC
2017
It's a quadratic over a quadratic, so use long division first and then the "remainder part" may be integrated using partial fraction decomposition (the quotient will just be a polynomial, which is easy to integrate).
I accidentally split the numerator before I did partial fractions so I had:

x + integral (5x-7)dx/(x^2 - 5x+6)

So I split it up into:

x + integral (5/2 * (2x-5) + 11/2)/(x^2 - 5x + 6)

x + 5/2 * integral (2x-5)/(x^2 - 5x+6) + 11/2 * integral (dx/(x-3)(x-2))

x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * integral (dx/(x-3)(x-2)) and then I did partial fractions on this

And got A = 1, B = -1

=> x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * integral (1/(x-3) - 1/(x-2) dx
= x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * ln|(x-3)/(x-2)| + C

But the answer is x - 3ln(x-2) +8ln(x-3) + C

Did I do a mistake in the working?
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,078
Gender
Male
HSC
N/A
I accidentally split the numerator before I did partial fractions so I had:

x + integral (5x-7)dx/(x^2 - 5x+6)

So I split it up into:

x + integral (5/2 * (2x-5) + 11/2)/(x^2 - 5x + 6)

x + 5/2 * integral (2x-5)/(x^2 - 5x+6) + 11/2 * integral (dx/(x-3)(x-2))

x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * integral (dx/(x-3)(x-2)) and then I did partial fractions on this

And got A = 1, B = -1

=> x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * integral (1/(x-3) - 1/(x-2) dx
= x + 5/2 * ln|x^2 - 5x + 6| + 11/2 * ln|(x-3)/(x-2)| + C

But the answer is x - 3ln(x-2) +8ln(x-3) + C

Did I do a mistake in the working?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top