What is the integral of (e^{6sqrt(t)})/(sqrt(t)) ?

The integral of (e^{6sqrt(t)})/(sqrt(t)) is 1/3 e^{6sqrt(t)}+C

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integral of (e^{6sqrt(t)})/(sqrt(t))

\int \frac{e ^{6 t}}{t} d t

\frac{1}{3} e^{6 t} + C

Solution steps

\int \frac{e ^{6 t}}{t} d t

Apply u-substitution

: \int \frac{e ^{u}}{3} d u

= \int \frac{e ^{u}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 6 t

= \frac{1}{3} e^{6 t}

Add a constant to the solution

= \frac{1}{3} e^{6 t} + C

\frac{d}{d x} (e^{(\frac{- k}{x})})

2 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

f (x) = \frac{1}{3 x}, at x = 9

x \to - 7 lim (\frac{7 - ∣ x ∣}{7 + x})

\int_{1}^{5} \frac{ln ( y )}{x y} d y

What is the integral of (e^{6sqrt(t)})/(sqrt(t)) ?
The integral of (e^{6sqrt(t)})/(sqrt(t)) is 1/3 e^{6sqrt(t)}+C