What is the integral of e^{1/2 t} ?

The integral of e^{1/2 t} is 2e^{1/2 t}+C

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integral of e^{1/2 t}

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

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

Solution steps

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

Apply u-substitution

: \int e^{u} \cdot 2 d u

= \int e^{u} \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 2 e^{u}

Substitute back

u = \frac{1}{2} t

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

Add a constant to the solution

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

7 t^{7} + 9

x^{- m}

\int \frac{1}{6 x - x ^{2}} d x

\frac{e ^{- π s}}{s ^{2} + 4}

\frac{d}{d x} (40 (x^{2} - 2 x))