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haiku11
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Homework Statement
v(t) = (-2 / (1 + 2t)) + 2
I must find the antiderivative s(t) such that s(0) = 5.
The Attempt at a Solution
u = 1 + 2t du/dt = 2 dt = du/2
√(-2 / (1 + 2t dt)) + 2
=√(-2u^-1 du/2) + 2
=√(-u^-1 du) + 2
= -ln|1+2t| + 2t + C
At t = 0
-ln(1 + 2(0)) + 2(0) + C = 5
-ln1 + C = 5
C = 5
s(t) = -ln|1+2t| + 2t + 5
I'm not sure if this is right, I derived it back and I got the original but there are so many things that are making me doubt myself. I have to do an antiderivative assignment without any prior knowledge so the only source I have is the internet which sometimes conflicts with itself. I read somewhere that if you have 1/x you get the special case of x^(-1 + 1) / (-1 + 1) in which case it becomes ln|x|. Is this true? Because http://answers.yahoo.com/question/index?qid=20090226170743AA3XucJ" says it's log x instead.
Also the dt is supposed to come right after each occurence of t right? Because that's what happens when you derive.
And finally (hopefully) when do I have to use substitution like I did in this question? Is it only in situations like log (1+x) or 2(2+x) or 123/(5+2x)? And I have to encompass everything in the bracket in the substitution right?
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