Hi all, Could someone please show me how to integrate: [ ( 1 + 4kut ) ^1/2 ] / 2kt dt I've tried everything, maybe I'm just being dim but I can't find a substitution that works and I can't do it by parts.
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Originally Posted by Ivanator27 Hi all, Could someone please show me how to integrate: [ ( 1 + 4kut ) ^1/2 ] / 2kt dt I've tried everything, maybe I'm just being dim but I can't find a substitution that works and I can't do it by parts. Let and the integral becomes You can now integrate this using partial fractions
I agree with that, however once: Which leads me to:
Last edited by Ivanator27; Apr 26th 2012 at 12:45 AM.
Let z^2=1+4kut SO 2zdz/dt=4ku So zdz/2ku z^2-1=4kut So 2kt=(z^2-1)/u Integral becomes 1/2k integral of z^2/(z^2-1) Now write z^2/(z^2-1)= 1+1/(z^2-1) and partial fraction 1/(z^2-1)
End of my 1st line should have been 2kt=(z^2-1)/2u
I don't see how this helps me, it leads to: Which is exactly what happens if you let Maybe I'm not understanding exactly what you were saying.
It was just an alternative substitution. You now need to write t in terms of z.
Ohhhh! Thanks so much! I have it know. =D ...Thanks again!
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