Question

integral as 0 approaches 1 (x-4)/x^2-5x+6 dx

Answer 1

factor the denominator and apply partial fractions

Answer 2

can you show me how to do that because i've tried and keep getting the wrong answer

Answer 3

how would you factor x^2-5x+6 ?

Answer 4

(x-3)(x-2)

Answer 5

i got up to integral from 0-1 6/x-2 - 5/x-3

Answer 6

Can you find A and B for this equation to be true?
\[\int\limits_{0}^{1}{\frac{x-4}{(x-2)(x-3)}}dx=\int\limits_{0}^{1}(\frac{A}{x-2}+\frac{B}{x-3})dx\]

Answer 7

i done all this . i just dont know how to integrate the ln

Answer 8

Can you write the integral?

Answer 9

If you mean this:

Answer 10

i got 6ln(x-2) - 5ln(x-3) } 1-0

Answer 11

Same principle i guess. Or you can write 6ln(x-2) as 6lnx/ln2=(6/ln2)*lnx where int lnx have been done.

Answer 12

i have no clue, can you help me out from top to bottom?

Answer 13

Can you write the integral you want to evaluate in equation mode first?