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

The integral of 1/(3^x) is -1/(ln(3)*3^x)+C

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integral of 1/(3^x)

integral

\frac{1}{3 ^{x}}

- \frac{1}{ln ( 3 ) \cdot 3 ^{x}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{ln ( 3 ) u ^{2}} d u

= \int \frac{1}{ln ( 3 ) u ^{2}} d u

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 3^{x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{e ^{x}}{( 2 + e ^{x} )}

e^{- x - 3}

r^{3}

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

sin (t) cos^{2} (t)