Question

calculus 2 - partial fractions

Answer 1

\[\frac{ 19x +1 }{ x^2 - 6x + 5 } = \frac{ 19x + 1 }{ (x- 5)(x - 1) }\]
we are doing the reverse step of adding fractions with different denominators.
we know that there exists A and B such that:\[\frac{ 19x + 1 }{ (x- 5)(x-1) } = \frac{ A }{ x - 5 } + \frac{ B }{ x - 1 }\]
we want to solve for A and B. so we add them again. when you have different denominators, you must put both numbers on common denominators:\[\frac{ A }{ x- 5 } + \frac{ B }{ x- 1 } = \frac{ A }{ x-5}\times\frac{ x-1 }{ x-1 } + \frac{ B }{ x-1 }\times\frac{ x-5 }{ x-5 } = \frac{ A(x-1) + B(x-5) }{ (x-1)(x-5) }\]\[=\frac{ Ax - A +Bx - 5B }{ (x-1)(x-5) }\]
now we know that:\[Ax - A + Bx - 5B = 19x + 1\]

do you know how to solve for A and B?

Answer 2

it becomes A(x-1) + B(x-5) = 19x + 1?

Answer 3

yes. but you can find real values for A and B that satisfy that equation

Answer 4

to solve it, isolate the terms:

Ax + Bx - A - 5B = 19x + 1
(A + B)x - (A + 5B) = (19)x + 1

so,
A + B = 19
and
A + 5B = -1

substitute B = 19 - A from the first of the two, into the second one
so the second one is:
A + 5(19 - A) = -1
A + 95 - 5A = -1
4A = 96
A = 24

plug A into substitution:
B = 19 - A = 19 - 24
B = -5

so, that means that:
\[\frac{ 19 + 1 }{ (x -5)(x-1) } = \frac{ 24 }{ x-5 } - \frac{ 5 }{ x - 1 }\]

this should be easy to integrate ^_^

Answer 5

24ln(x-5) -5ln(x-1

Answer 6

yes

Answer 7

any absolute values?

Answer 8

yes

Answer 9

Thank you

Answer 10

Just to add this here next to @Euler271 flawless approach and explanation, once you arrived at:
\[A(x-1) + B(x-5) = 19x + 1 \]
You could also set \(x=1 \) which will give you \(B=-4\) and setting \(x=5\) will give you \(A=25\). Just to demonstrate an alternative approach (which can get vastly complicated for more than 3 unknowns though)

Answer 11

happy to help ^_^

and yes, that is a good way to do it @Spacelimbus