What is the integral of 3^{2x}e^x ?

The integral of 3^{2x}e^x is ((9e)^x)/(ln(9e))+C

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integral of 3^{2x}e^x

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

\frac{( 9 e ) ^{x}}{ln ( 9 e )} + C

Solution steps

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

3^{2 x} = 9^{x}

= \int 9^{x} e^{x} d x

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

9^{x} e^{x} = (9 e)^{x}

= \int (9 e)^{x} d x

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

= \frac{( 9 e ) ^{x}}{ln ( 9 e )}

Add a constant to the solution

= \frac{( 9 e ) ^{x}}{ln ( 9 e )} + C

x \to - 4 lim (∣ x ∣)

\frac{d}{d x} (ln (x^{3} + 4))

h (x) = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

y^{^{'}} = 2 cos (5 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{2}

What is the integral of 3^{2x}e^x ?
The integral of 3^{2x}e^x is ((9e)^x)/(ln(9e))+C