Question

integral of dx/(Sqrt(-x^2+8x-12))

Answer 1

-2x^3/3+8x^2/2-12x

Answer 2

which can be simplified:

-2x^3/3 + 4x^2-12x

Answer 3

sqrt goes to x^(1/2)

Answer 4

bit lost on the sqrt its been a while. can someone else help?

Answer 5

1/x goes to ln x

Answer 6

\[\int\frac{dx}{\sqrt{-x^2+8x-12}}\]
Complete the square in the denominator:
\[-x^2+8x-12\\
-\left(x^2-8x\right)-12\\
-\left(x^2-8x+16-16\right)-12\\
-\left((x-4)^2-16\right)-12\\
-(x-4)^2+16-12\\
4-(x-4)^2\]
So now you have
\[\int\frac{dx}{\sqrt{4-(x-4)^2}}\]
Now make a trigonometric substitution. Let
\[x-4=2\sin u\\
dx=2\cos u~du\]
\[\int\frac{2\cos u}{\sqrt{4-(2\cos u)^2}}~du\]

Answer 7

Sorry, that last integral should be
\[\int\frac{2\cos u}{\sqrt{4-(2\color{red}{\sin u})^2}}~du\]

Answer 8

why make a sub with trig functions?

Answer 9

What would you do instead? The procedure you suggested doesn't work.