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

The integral of ((a^xln(a)))/((a^x+1)) is ln|a^x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((a^xln(a)))/((a^x+1))

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

ln ∣ a^{x} + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln (a) \frac{1}{ln ( a )} ln ∣ u ∣

Substitute back

u = a^{x} + 1

= ln (a) \frac{1}{ln ( a )} ln ∣ a^{x} + 1 ∣

Simplify

ln (a) \frac{1}{ln ( a )} ln ∣ a^{x} + 1 ∣ : ln ∣ a^{x} + 1 ∣

= ln ∣ a^{x} + 1 ∣

Add a constant to the solution

= ln ∣ a^{x} + 1 ∣ + C

\int \frac{8}{( x ^{2} - 4 )} d x

\int \frac{( e ^{x} e ^{x} - 1 )}{e ^{x} + 8} d x

\int \frac{1}{( y )} d y

\int 0.09 e^{0.04 x} d x

\int \frac{( x ^{3} )}{( x ^{2} + 4 ) ^{(\frac{1}{2})}} d x

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