What is #f(x) = int 1/(x-3)-x/(x-2) dx# if #f(-1)=6 #?

1 Answer
Alexander L.
Feb 7, 2016

#-x+ln(x-3)-2ln(x-2)+2+c#

Explanation:

#int1/(x-3)-x/(x-2)dx#
#ln(x-3)-intx/(x-2)dx#
..............................................................................................................................
#intx/(x-2)dx=intx(x-2)^-1dx#
Via integration by parts
#xln(x-2)-intln(x-2)dx#

Let #y=ln(x-2)# so #e^y=x-2# and #e^ydy=dx#
#xln(x-2)-intln(x-2)dx=xln(x-2)-intye^ydy#
Integrate by parts again.
#xln(x-2)-intln(x-2)dx=xln(x-2)-ye^y+e^y#
#xln(x-2)-(x-2)ln(x-2)+x-2#
..............................................................................................................................
#int1/(x-3)-x/(x-2)dx#
#ln(x-3)-intx/(x-2)dx#
#ln(x-3)-xln(x-2)+(x-2)ln(x-2)-x+2#
Simplify:
#f(x)=ln(x-3)-2ln(x-2)-x+2+C#
Substitute #f(-1)=6#

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