What is the integral of e^y2^{e^y}3^{e^y} ?

The integral of e^y2^{e^y}3^{e^y} is (6^{e^y})/(ln(6))+C

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integral of e^y2^{e^y}3^{e^y}

\int e^{y} 2^{e^{y}} 3^{e^{y}} d y

\frac{6 ^{e^{y}}}{ln ( 6 )} + C

Solution steps

\int e^{y} \cdot 2^{e^{y}} \cdot 3^{e^{y}} d y

Apply u-substitution

: \int 6^{u} d u

= \int 6^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{6 ^{u}}{ln ( 6 )}

Substitute back

u = e^{y}

= \frac{6 ^{e^{y}}}{ln ( 6 )}

Add a constant to the solution

= \frac{6 ^{e^{y}}}{ln ( 6 )} + C

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

x^{2} - 6 x + 3

f (x) = ln (x^{2} - 3 x + 1), (3, 0)

\int_{- 5}^{0} (\frac{y}{5} + y + 6) d y

\frac{d y}{d x}, x y = 84

What is the integral of e^y2^{e^y}3^{e^y} ?
The integral of e^y2^{e^y}3^{e^y} is (6^{e^y})/(ln(6))+C