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−3dx+bx−2dx
ax + bx = 1ax+bx=1
a + b = 1a+b=1
-2a - 3b = -4−2a−3b=−4
Solving it out, we get a = -1a=−1 and b = 2b=2
int(-1)/(x-3)dx+int2/(x-2)dx∫−1x−3dx+∫2x−2dx
-int1/(x-3)dx+2int1/(x-2)dx−∫1x−3dx+2∫1x−2dx
Integral this -int1/(x-3)dx+int1/(x-2)dx+int1/(x-2)dx−∫1x−3dx+∫1x−2dx+∫1x−2dx
to get this -ln|x-3| + 2ln|x-2|−ln|x−3|+2ln|x−2|
Now to integral [-ln|x-3| + 2ln|x-2|]_0^1[−ln|x−3|+2ln|x−2|]10
(-ln|1-3| + 2ln|1-2|) - (-ln|0-3| + 2ln|0-2|)(−ln|1−3|+2ln|1−2|)−(−ln|0−3|+2ln|0−2|)
=(-ln|-2| + 2ln|-1|) - (-ln|-3| + 2ln|-2|)(−ln|−2|+2ln|−1|)−(−ln|−3|+2ln|−2|)
=(-ln2 + ln1) - (-ln3 + 2ln2)(−ln2+ln1)−(−ln3+2ln2)
=-ln2 - ln4/3−ln2−ln43
=-ln8/3−ln83
=-.981−.981
