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

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

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integral of (3^x)/(3-3^x)

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{ln ( 3 )} ln ∣ u ∣

Substitute back

u = 3 - 3^{x}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{ln ( x )}{sin ( x )})

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

\int \frac{7}{x ^{2}} d x

\int 20 x^{3} d x

\frac{d Q}{d t} = 0.03 (1 - Q)^{\frac{2}{3}}

What is the integral of (3^x)/(3-3^x) ?
The integral of (3^x)/(3-3^x) is -1/(ln(3))ln|3-3^x|+C