What is the integral of 9e^{x/4} ?

The integral of 9e^{x/4} is 36e^{x/4}+C

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integral of 9e^{x/4}

\int 9 e^{\frac{x}{4}} d x

36 e^{\frac{x}{4}} + C

Solution steps

\int 9 e^{\frac{x}{4}} d x

Take the constant out:

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

= 9 \cdot \int e^{\frac{x}{4}} d x

Apply u-substitution

: \int e^{u} \cdot 4 d u

= 9 \cdot \int e^{u} \cdot 4 d u

Take the constant out:

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

= 9 \cdot 4 \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot 4 e^{u}

Substitute back

u = \frac{x}{4}

= 9 \cdot 4 e^{\frac{x}{4}}

Simplify

= 36 e^{\frac{x}{4}}

Add a constant to the solution

= 36 e^{\frac{x}{4}} + C

\int_{- \infty}^{0} x e^{- x} d x

y = 5 x, (5, 5)

\int 3 x^{2} e^{2 x^{3}} d x

x - x^{2}

(4 t)^{^{'}}