What is the integral of 8e^{4x} ?

The integral of 8e^{4x} is 2e^{4x}+C

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integral of 8e^{4x}

\int 8 e^{4 x} d x

2 e^{4 x} + C

Solution steps

\int 8 e^{4 x} d x

Take the constant out:

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

= 8 \cdot \int e^{4 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= 8 \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= 8 \cdot \frac{1}{4} e^{4 x}

Simplify

8 \cdot \frac{1}{4} e^{4 x} : 2 e^{4 x}

= 2 e^{4 x}

Add a constant to the solution

= 2 e^{4 x} + C

y = 6 x

\int_{- 1}^{1} \frac{2}{x ^{2} + 1} d x

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 8 x)^{12})

f (x) = - cos (5 x)

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