• Nov 27th 2009, 12:10 PM
shmayer
I am suppose to evaluate the indefinite integral of
dx / square root of (1 - x^2 +2x)..
I first broke down 1-2^x + 2x into 2-(x-1)^2
I then took the antiderivative of dx/(2-(x-1)^2) and came up with the answer of [sin^-1(x-1) / 2] + C
Can someone please tell me if this is correct, or what i can do to make it right? I took the derivavtive and it seems i might have made a small error somewhere in finding the antiderivative but im not soo sure where...

Also, if someone can help me solve this question, thatd be great, thanks!!(Rofl)
Compute the integral from -1 ->3 of the square root of (6-2x^2 +4x)dx .
I can't find the antiderivative of this function...and substitution doesnt seem to work...
• Nov 27th 2009, 01:39 PM
skeeter
Quote:

Originally Posted by shmayer
I am suppose to evaluate the indefinite integral of
dx / square root of (1 - x^2 +2x)..
I first broke down 1-2^x + 2x into 2-(x-1)^2
I then took the antiderivative of dx/(2-(x-1)^2) and came up with the answer of [sin^-1(x-1) / 2] + C
Can someone please tell me if this is correct, or what i can do to make it right? I took the derivavtive and it seems i might have made a small error somewhere in finding the antiderivative but im not soo sure where...

Compute the integral from -1 ->3 of the square root of (6-2x^2 +4x)dx .
I can't find the antiderivative of this function...and substitution doesnt seem to work...

I thought I saw these two exact problems before ...

http://www.mathhelpforum.com/math-he...gral-help.html

... did you even bother to look?
• Nov 27th 2009, 01:42 PM
Abu-Khalil
$\int\frac{dx}{\sqrt{1-x^2+2x}}=\int\frac{dx}{\sqrt{2-(x-1)^2}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{1-\left(\frac{x-1}{\sqrt{2}}\right)^2}}=\arcsin \left(\frac{x-1}{\sqrt{2}}\right)+C$

PD: Ups. Late post.