Question

\[\int\limits (x+5)/(\sqrt{9-(x-3)^2})dx\]

Answer 1

Okay, this may be weird but do this. Make the substitution:
\[\rho =x-3; \rho+3=\rho; d \rho=d x\]
Then we get:
\[\int\limits \frac{(\rho+3)+5}{\sqrt{9-\rho^2}}d \rho=\int\limits \frac{\rho}{\sqrt{9-\rho^2}}d \rho+8 \int\limits \frac{d \rho}{\sqrt{9-\rho^2}}\]
Can you finish it from there? The right most integral is arcsin i believe and the left one is another u-sub.

Answer 2

wait, p+3=p? is that allowed?

Answer 3

Oh sorry, that should be p+3=x ***

Answer 4

oh. It makes sense. I'm gonna try it

Answer 5

I just added the 3 over x-3=p implies x=p+3.

Answer 6

yep :)

Answer 7

Got it! Thanks!
\[-\sqrt{6x-x^2} + 8/3 \arcsin [(x-3)/3]\]

Answer 8

Nice!