Question

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

Answer 1

`-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`