Integral homework problem question

Could someone explain how my teacher got from step 1 to step 2. I am very new to integrals so please don't skip key things and assume I know them.
The integral has X at the top of the "S" and 0 at the bottom of the "S"
Step 1: [dX/(A-BX)] into Step 2: (-1/B)[ln(A-BX)-ln(A)]
I don't get how he went from step 1 to step 2. THANK YOU!

ok i see how to help

first you want to use substitution

the part on the bottom is a bit complex ... lets change it

u = A - BX

now take the derivative of this

du/dx = -B

-du/B = dx

==========

thats step one

now we can change the integral

substitute the (-du/B) for the dx

and u for the (A - BX)

integral -(du/B)/u

integral -(1/B)du/u

so we can take the -(1/B) out of the integral

-(1/B) integral du/u

the integral of du/u = ln u

this you just have to know

because d(lnu)/du = 1/u

I cant hellp you there you just have to know it

so we have the solution

-(1/B) integral du/u

-(1/B) integral of du/u = -(1/B)ln u

now we change the ln u back to the value it was substituted for

u = A - BX

-(1/B) ln(A-BX)

now remember this has to be evaluated at the two points we were given

the point on the top of the S or integral sign was X = x

the point on the bottom o f the integral sign was X = 0

I put the X for the variable in the integral and the little x for the value of the endpoint on the integral sign

when you evaluate at the end points you do the top value first then subtract it from the bottom point..
...........................|X = x
-(1/B) ln(A-BX) |
..........................| X = 0

becomes

-(1/B)* (ln(A -Bx) - ln(A - B*0))

obviously 0*B = 0

-(1/B)*(ln(A - Bx) - ln(A))

that is how you get that answer

I hope I helped its hard to show the integral sign in this text mode..