Question

S dx/sqrt(3+2x-x^2)

Answer 1

when you see sqrt in the denominator perhaps the integral will be an inverse trig function
http://www.math.brown.edu/UTRA/trigderivs.html#inverses

Try to "complete the square" on -x^2 + 2x

Answer 2

i got a square of 4-(x-1)^2

Answer 3

after simplifying the sqrt i got dx/-x+1
which becames -ln|-x+1|+c

Answer 4

you mean square root:
\[ \int \frac{dx}{\sqrt{4 - (x-1)^2} }\]
factor 4 from out of the square root:
\[ \int \frac{dx}{2\sqrt{1 - \frac{(x-1)^2}{4}} }\\
\int \frac{dx}{2\sqrt{1 - \left(\frac{x-1}{2}\right)^2} }
\]

Answer 5

let u = (x-1)/2
find du in terms of dx, solve for dx and sub in
what do you get ?

Answer 6

i split the square root \[\frac{ dx }{\sqrt{4-(x-1)^2} } \to \frac{ dx }{ 2-x-1 }\] is that wrong?

Answer 7

unfortunately , the square root of 4 - (x-1)^2 is not what you wrote
2-x-1 (or more simply 1-x ) squared is (1-2x+x^2)
on the other hand 4-(x-1)^2 is 4 - x^2 +2x -1 = -x^2+2x+3

so you have to keep the radical sign.

But if you use the substitution u = (x-1)/2

you will get an expression that can be integrated using the formula for the inverse sign.

Answer 8

***inverse sine