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

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

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

integral

\frac{1}{( a ^{x} + 1 )}

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

Solution steps

\int \frac{1}{a ^{x} + 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{( u - 1 ) u} : \frac{1}{u - 1} - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{ln ( a )} (ln ∣ u - 1 ∣ - ln ∣ u ∣)

Substitute back

u = a^{x} + 1

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

Simplify

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

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

Add a constant to the solution

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

(x + 2)^{2}

e^{\frac{- 2}{x}}

\frac{[ ln ( ( 2 x ) ^{2} ) ]}{x}

(x^{2} + 1) cos^{2} (x)

\frac{r ^{2}}{1 + r ^{2}}

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