What is the integral of 1/2 e^{2x+1} ?

The integral of 1/2 e^{2x+1} is 1/4 e^{2x+1}+C

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integral of 1/2 e^{2x+1}

\int \frac{1}{2} e^{2 x + 1} d x

\frac{1}{4} e^{2 x + 1} + C

Solution steps

\int \frac{1}{2} e^{2 x + 1} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{2 x + 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= \frac{1}{2} \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{1}{2} e^{2 x + 1}

Simplify

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

= \frac{1}{4} e^{2 x + 1}

Add a constant to the solution

= \frac{1}{4} e^{2 x + 1} + C

x x

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

y \to - 2 lim (y^{3} (5 - 3 y^{2}))

y^{^{'}} = \frac{y}{x} + e^{- \frac{y}{x}}

\int_{0}^{2} k (2 - x)

What is the integral of 1/2 e^{2x+1} ?
The integral of 1/2 e^{2x+1} is 1/4 e^{2x+1}+C