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

The integral of (a^xln(a))/(1+a^{2x)} is arctan(a^x)+C

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

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

arctan (a^{x}) + C

Solution steps

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

Take the constant out:

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

= ln (a) \cdot \int \frac{a ^{x}}{1 + a ^{2 x}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= ln (a) \frac{1}{2 ln ( a )} \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

= ln (a) \frac{1}{2 ln ( a )} \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)

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

Substitute back

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

Simplify

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

= arctan (a^{x})

Add a constant to the solution

= arctan (a^{x}) + C

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What is the integral of (a^xln(a))/(1+a^{2x)} ?
The integral of (a^xln(a))/(1+a^{2x)} is arctan(a^x)+C