Question

Question #8e17c

Answer 1

`ln|(x - 2)/(x - 3)| + ln|x-2|`

Explanation:

Right off the bat, we can see that we will need to use impartial fraction integrals.

We set it as `inta/(x-3)dx + b/(x-2)dx`
`ax + bx = 1`
`a + b = 1`
`-2a - 3b = -4`

Solving it out, we get `a = -1` and `b = 2`

`int(-1)/(x-3)dx+int2/(x-2)dx`
`-int1/(x-3)dx+2int1/(x-2)dx`
Integral this `-int1/(x-3)dx+int1/(x-2)dx+int1/(x-2)dx`

to get this `-ln|x-3| + 2ln|x-2|`

Now to integral `[-ln|x-3| + 2ln|x-2|]_0^1`

`(-ln|1-3| + 2ln|1-2|) - (-ln|0-3| + 2ln|0-2|)`
=`(-ln|-2| + 2ln|-1|) - (-ln|-3| + 2ln|-2|)`
=`(-ln2 + ln1) - (-ln3 + 2ln2)`
=`-ln2 - ln4/3`
=`-ln8/3`
=`-.981`