Question

I'm having trouble taking the indefinite integral of
-1/(x^2-6x+9) dx
Why isn't the answer -lnabs(x-3)+c

Answer 1

Hmm well lets see

Firstly lets factor that bottom part

\[-\int\limits^{}_{}\frac{1}{(x - 3)^2}dx \]

Now make u = (x - 3)
du = dx

\[ -\int\limits^{}_{}\frac{1}{u^2}du \]

The integral of 1/u^2 = -1/u + C so

\[\ -\frac{-1}{u} + c \]
\[\ \frac{1}{u} + C \]

Plug back in u = (x - 3)

\[\frac{1}{x - 3}+ C\]

Answer 2

oooh ok thank you so much!

Answer 3

No problem!