What is the integral of 1/4 e^{4x+1} ?

The integral of 1/4 e^{4x+1} is 1/16 e^{4x+1}+C

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integral of 1/4 e^{4x+1}

\int \frac{1}{4} e^{4 x + 1} d x

\frac{1}{16} e^{4 x + 1} + C

Solution steps

\int \frac{1}{4} e^{4 x + 1} d x

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{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

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

Use the common integral:

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

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

Substitute back

u = 4 x + 1

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

Simplify

\frac{1}{4} \cdot \frac{1}{4} e^{4 x + 1} : \frac{1}{16} e^{4 x + 1}

= \frac{1}{16} e^{4 x + 1}

Add a constant to the solution

= \frac{1}{16} e^{4 x + 1} + C

- x^{2} + 80 x - 600

sin^{2} (π x)

\int \frac{24 x - 2 x ^{3}}{x ^{4} - 16} d x

\frac{d}{d x} (\frac{8}{x} - \frac{4}{x})

\frac{\partial}{\partial y} (yz - ln (x + z))

What is the integral of 1/4 e^{4x+1} ?
The integral of 1/4 e^{4x+1} is 1/16 e^{4x+1}+C