What is the integral of (3^x)/(1+9^x) ?

The integral of (3^x)/(1+9^x) is (arctan(3^x))/(ln(3))+C

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integral of (3^x)/(1+9^x)

\int \frac{3 ^{x}}{1 + 9 ^{x}} d x

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

Solution steps

\int \frac{3 ^{x}}{1 + 9 ^{x}} d x

Apply u-substitution

: \int \frac{1}{2 ln ( 3 ) \cdot 3 ^{\frac{l n ( u - 1 )}{2 l n ( 3 )}} u} d u

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

Apply u-substitution

: \int \frac{2}{v ^{2} + 1} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2 ln ( 3 )} \cdot 2 arctan (v)

Substitute back

= \frac{1}{2 ln ( 3 )} \cdot 2 arctan (3^{\frac{l n ( 1 + 9 ^{x} - 1 )}{2 l n ( 3 )}})

Simplify

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

= \frac{arctan ( 3 ^{x} )}{ln ( 3 )}

Add a constant to the solution

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

- \frac{12}{x ^{5}}

f (x) = - 3 x^{2} - 4 x

\frac{d}{d y} (\frac{1 - cos ( y )}{1 - cos ( 2 y )})

\frac{150 t}{t + 3}

\int t 9 t^{2} + 4 d t

What is the integral of (3^x)/(1+9^x) ?
The integral of (3^x)/(1+9^x) is (arctan(3^x))/(ln(3))+C