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anonymous
 3 years ago
Find the indefinite integral of ((1/2t)(2)^(1/2)e^t)dt
anonymous
 3 years ago
Find the indefinite integral of ((1/2t)(2)^(1/2)e^t)dt

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large\int\frac12t2^{\frac12e^t}dt\]This?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, the first t is in bottom with the 2, and the second 2 is under a radical by itself and e^t is another variable not an entire exponent

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \int \frac{1}{2t}dt 2^{1/2} \int e^t dt\] If that's not what you mean I suggest you, trying to \( \LaTeX\) it out yourself, so we all talk about the same problem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's almost right just include e^t after the square root of 2 and take off dt till the end

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\infty}^{\infty}[\frac{ 1 }{ 2t }\sqrt{2}e^t]dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that your question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Spacelimbus is right you need to tell him the right question so he can help you to provide you with the right solution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits[\frac{ 1 }{ 2t }\sqrt{2}e^t]dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry this is an indefinite integral i think you are asking

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if this is the right question then the answer would be

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{ 1 }{ 2 }\ln(t)2^\frac{ 1 }{2 } e^t+c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0otherwise help us to help you by providing the right question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes that's the right question! But how you solve it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Forget it, I got it, thank you
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