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

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

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integral of 4e^{4x}

\int 4 e^{4 x} d x

e^{4 x} + C

Solution steps

\int 4 e^{4 x} d x

Take the constant out:

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

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

Apply u-substitution

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

= 4 \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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

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

= e^{4 x}

Add a constant to the solution

= e^{4 x} + C

\int \frac{x}{x ^{3} + x ^{2} + 4 x + 4} d x

t - 2

\frac{d y}{d x} = y (x y^{2} - 1)

\int (tan (x) + sec (x))^{2} d x

x \to \infty lim (\frac{x ^{3} - 4 x ^{2} + x + 6}{x + 1})