How can i integrate Ln (1-x/5 )?

Answer 1

∫ln (1-x/5) dx

Proceed with integration by parts: u=ln (1-x/5), v=x, du=1/(1-x/5) * -1/5 dx, dv=dx. The formula is ∫u dv = uv - ∫v du, so:

x ln (1-x/5) + ∫1/5 * x/(1-x/5) dx

Simplifying:

x ln (1-x/5) + ∫x/(5-x) dx

Rewriting the numerator:

x ln (1-x/5) + ∫(x-5)/(5-x) + 5/(5-x) dx

Simplifying:

x ln (1-x/5) - ∫1 dx + ∫5/(5-x) dx

Now integrating:

x ln (1-x/5) - x - 5 ln (5-x) + C

And we are done.

Edit: corrected egregious error -- I had forgotten to include the x in the integral when applying integration by parts. Interestingly, the thumbs up I got (or one of them, if I get any after this edit) was given despite this error. I really wish people would actually verify the correctness of an answer before giving thumbs up and not simply assume that it's right simply because I have a top contributor tag under my name.

Answer 2

Using integration by parts:

int( Ln (1-x/5))dx
note that int means 'integral of'

solution:
let u = ln(1-x/5)
du = (1/(1-x/5))*(-1/5)
dv = dx
v = x

Using the formula: integrand = uv - int(vdu)

integrand = xln(1-x/5) + (int (x/(5-x)))
= xln(1-x/5) + int ((x-2x+5-5)/(5-x))
= xln(1-x/5) + int ((-x+5)/(5-x) -5/(5-x))
= xln(1-x/5) + int ( 1 - 5/(5-x))
= xln(1-x/5) + x + 5ln(5-x)


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xln(1-x/5) + x + 5ln(5-x)
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Answer 3

Ewww integration I almost failed high school because of that lol.

Answer 4

Hehe......Ummmm.........6?!?