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

1 Answer
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
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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
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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