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

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

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integral of 3^{2-x}

\int 3^{2 - x} d x

- \frac{3 ^{2 - x}}{ln ( 3 )} + C

Solution steps

\int 3^{2 - x} d x

Apply u-substitution

: \int - 3^{u} d u

= \int - 3^{u} d u

Take the constant out:

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

= - \int 3^{u} d u

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

= - \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = 2 - x

= - \frac{3 ^{2 - x}}{ln ( 3 )}

Add a constant to the solution

= - \frac{3 ^{2 - x}}{ln ( 3 )} + C

\int \frac{1}{x ^{2} - 2 x - 63} d x

\int_{\frac{8}{3}}^{4} 3 t - 8 d t

\frac{d}{d x} (ln (\frac{a x + 1}{a x - 1}))

f (x) = 3 x + \frac{3}{x}

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

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