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

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

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

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

\frac{1}{2 ln ( 3 )} ln ∣ 1 + 9^{x} ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2 ln ( 3 )} ln ∣ u ∣

Substitute back

u = 1 + 3^{2 x}

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

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

= \frac{1}{2 ln ( 3 )} ln ∣ 1 + 9^{x} ∣

Add a constant to the solution

= \frac{1}{2 ln ( 3 )} ln ∣ 1 + 9^{x} ∣ + C

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