What is the integral of 6(e^{2x})/8 ?

The integral of 6(e^{2x})/8 is 3/8 e^{2x}+C

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integral of 6(e^{2x})/8

\int 6 \frac{e ^{2 x}}{8} d x

\frac{3}{8} e^{2 x} + C

Solution steps

\int 6 \cdot \frac{e ^{2 x}}{8} d x

Take the constant out:

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

= 6 \cdot \frac{1}{8} \cdot \int e^{2 x} d x

6 \cdot \frac{1}{8} = \frac{3}{4}

= \frac{3}{4} \cdot \int e^{2 x} d x

Apply u-substitution

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

= \frac{3}{4} \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{3}{4} \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{3}{4} \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x

= \frac{3}{4} \cdot \frac{1}{2} e^{2 x}

Simplify

\frac{3}{4} \cdot \frac{1}{2} e^{2 x} : \frac{3}{8} e^{2 x}

= \frac{3}{8} e^{2 x}

Add a constant to the solution

= \frac{3}{8} e^{2 x} + C

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

\int \frac{10}{1 + 4 x ^{2}} d x

\int \frac{x ^{2} + 6}{x ^{3} + x} d x

\frac{d}{d x} (\frac{5}{( 2 x - 3 ) ^{4}})

(0, 5), (10, 0)

What is the integral of 6(e^{2x})/8 ?
The integral of 6(e^{2x})/8 is 3/8 e^{2x}+C