First, separate it into the format:
A/(x-5) + B/(x-3)Ax−5+Bx−3
Now cross-multiply and simplify:
(A(x-3) + B(x-5))/((x-5)(x-3))A(x−3)+B(x−5)(x−5)(x−3)
= (Ax - 3A + Bx - 5B)/((x-5)(x-3))=Ax−3A+Bx−5B(x−5)(x−3)
= ((A + B)x - 3A - 5B)/((x-5)(x-3))=(A+B)x−3A−5B(x−5)(x−3)
Notice there's an xx term and an x^0x0 term (constants). We can equate them back to what the numerator originally was. There was no xx term originally, so:
A + B = 0A+B=0
A = -BA=−B
-3A - 5B = 2−3A−5B=2
=> 3B - 5B = 2⇒3B−5B=2
-2B = 2−2B=2
=> B = -1⇒B=−1
=> A = 1⇒A=1
You're basically done. I was taught that int1/xdx = ln|x| + C∫1xdx=ln|x|+C... so:
= int 1/(x-5) - 1/(x-3)dx=∫1x−5−1x−3dx
= ln|x-5| - ln|x-3| + C=ln|x−5|−ln|x−3|+C
Wolfram Alpha doesn't seem to agree with me about absolute values here compared to here, so let's just try differentiating this to see what we get.
[(df(x))/(dx)] " at " x > 0 = 1/(x-5) - 1/(x-3)[df(x)dx] at x>0=1x−5−1x−3
= ((x-3) - (x-5))/((x-5)(x-3))=(x−3)−(x−5)(x−5)(x−3)
= (x-3 - x+5)/((x-5)(x-3))=x−3−x+5(x−5)(x−3)
= 2/((x-5)(x-3))=2(x−5)(x−3)
...Yup, it works. "*shrug*"*shrug*
