How do you integrate #2ln(x-5)#?

Question

1415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-05T21:34:21

#2int 1*ln(x-5)dx#

You can integrate this using integration by parts. The general formula for integration by parts is:

#uv - intvdu#

Now for w-substitution, let
#x-5 = w#
#x = w+5#
#dw = dx#

#2intln(x-5)dx = 2intlnwdw#

For integration by parts, let
#u = lnw#
#dv = 1dw#
#v = w#
#du = 1/wdw#

#= 2[wln|w| - intw/wdw]#

#= 2[wln|w| - w + C]#

#= 2[(x-5)ln|x-5| - (x-5) + C]#

#= 2[xln|x-5| - 5ln|x-5| - x + 5 + C]#

But #5# is a constant, so it gets embedded into #C#. Multiplying #C# by #2# still gives #C#.

#= 2[xln|x-5| - 5ln|x-5| - x] + C#