What is the integral of y=2e^{4x-3}6 ?

The integral of y=2e^{4x-3}6 is 3e^{4x-3}+C

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integral of y=2e^{4x-3}6

integral

y = 2 e^{4 x - 3} 6

3 e^{4 x - 3} + C

Solution steps

\int 2 e^{4 x - 3} \cdot 6 d x

Take the constant out:

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

= 2 \cdot 6 \cdot \int e^{4 x - 3} d x

Simplify

= 12 \cdot \int e^{4 x - 3} d x

Apply u-substitution

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

= 12 \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

= 12 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 12 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x - 3

= 12 \cdot \frac{1}{4} e^{4 x - 3}

Simplify

12 \cdot \frac{1}{4} e^{4 x - 3} : 3 e^{4 x - 3}

= 3 e^{4 x - 3}

Add a constant to the solution

= 3 e^{4 x - 3} + C

lo g_{10} (x^{2})

\frac{( x ^{5} + 1 )}{( x - 1 )}

\frac{e ^{x^{2}}}{x}

\frac{x ^{2} - 12 x + 13}{x}

\frac{( x + 3 )}{( x - 1 )}

What is the integral of y=2e^{4x-3}6 ?
The integral of y=2e^{4x-3}6 is 3e^{4x-3}+C