What is the integral of 1/(3^{2x)} ?

The integral of 1/(3^{2x)} is -(9^{-x})/(2ln(3))+C

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integral of 1/(3^{2x)}

\int \frac{1}{3 ^{2 x}} d x

- \frac{9 ^{- x}}{2 ln ( 3 )} + C

Solution steps

\int \frac{1}{3 ^{2 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{3 ^{2 x}} = 3^{- 2 x}

= \int 3^{- 2 x} d x

Apply u-substitution

: \int - \frac{1}{2} \cdot 3^{u} d u

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

Take the constant out:

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

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

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

= - \frac{1}{2} \cdot \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = - 2 x

= - \frac{1}{2} \cdot \frac{3 ^{- 2 x}}{ln ( 3 )}

Simplify

- \frac{1}{2} \cdot \frac{3 ^{- 2 x}}{ln ( 3 )} : - \frac{9 ^{- x}}{2 ln ( 3 )}

= - \frac{9 ^{- x}}{2 ln ( 3 )}

Add a constant to the solution

= - \frac{9 ^{- x}}{2 ln ( 3 )} + C

\frac{d}{d x} (- \frac{1}{2 x ^{\frac{3}{2}}})

x \to 0 lim (tan^{x} (2 x))

y = 2 x^{2} 25 - x^{2}

y^{^{'}} = x y - y

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

What is the integral of 1/(3^{2x)} ?
The integral of 1/(3^{2x)} is -(9^{-x})/(2ln(3))+C