What is the integral of (2t)^{-1/2} ?

The integral of (2t)^{-1/2} is 2^{1/2}t^{1/2}+C

integral of (2t)^{-1/2}

\int (2 t)^{- \frac{1}{2}} d t

2^{\frac{1}{2}} t^{\frac{1}{2}} + C

Solution steps

\int (2 t)^{- \frac{1}{2}} d t

Simplify

(2 t)^{- \frac{1}{2}} : \frac{1}{2 ^{\frac{1}{2}} t ^{\frac{1}{2}}}

= \int \frac{1}{2 ^{\frac{1}{2}} t ^{\frac{1}{2}}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 ^{\frac{1}{2}}} \cdot \int \frac{1}{t ^{\frac{1}{2}}} d t

Apply the Power Rule

= \frac{1}{2 ^{\frac{1}{2}}} \cdot 2 t^{\frac{1}{2}}

Simplify

\frac{1}{2 ^{\frac{1}{2}}} \cdot 2 t^{\frac{1}{2}} : 2^{\frac{1}{2}} t^{\frac{1}{2}}

= 2^{\frac{1}{2}} t^{\frac{1}{2}}

Add a constant to the solution

= 2^{\frac{1}{2}} t^{\frac{1}{2}} + C

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What is the integral of (2t)^{-1/2} ?
The integral of (2t)^{-1/2} is 2^{1/2}t^{1/2}+C