# Can someone explain to me why this is with finding an integral

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Alright, I know how to find the integral of a regular function, But an inverse, is throwing me off

Can someone explain why 1/Sqrt(7t)dt is 2sqrt(t/7)

I'm probably missing something, but why does the variable end up as a numerator?

I know the inverse is 1/(r+1)x^(r+1)

So I work it out as
1/(-.5+1) which goes out to 2

then the rest
(7t)^-1/2+1
(7t)^1/2
(what am I missing here)
=2sqrt(t/7)+C

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Ray Vickson
Homework Helper
Dearly Missed
Alright, I know how to find the integral of a regular function, But an inverse, is throwing me off

Can someone explain why 1/Sqrt(7t)dt is 2sqrt(t/7)

I'm probably missing something, but why does the variable end up as a numerator?

I know the inverse is 1/(r+1)x^(r+1)

So I work it out as
1/(-.5+1) which goes out to 2

then the rest
(7t)^-1/2+1
(7t)^1/2
(what am I missing here)
=2sqrt(t/7)+C
So, you are saying that
$$\int \frac{1}{\sqrt{7t}} \, dt = 2 \sqrt{t/7} + C$$
You can check by taking the derivative of your answer with respect to t and see if it gives you back the integrand. That is something you should always do, and when you do it you will have answered your own question.

RUber
Homework Helper
It always helps me to think of