What is the integral of 1/(10^x+1) ?

The integral of 1/(10^x+1) is -1/(ln(10))ln|10^x+1|+x+C

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integral of 1/(10^x+1)

\int \frac{1}{1 0 ^{x} + 1} d x

- \frac{1}{ln ( 10 )} ln ∣ 1 0^{x} + 1 ∣ + x + C

Solution steps

\int \frac{1}{1 0 ^{x} + 1} d x

\frac{1}{1 0 ^{x} + 1} = - \frac{1 0 ^{x}}{1 0 ^{x} + 1} + 1

= \int - \frac{1 0 ^{x}}{1 0 ^{x} + 1} + 1 d x

Apply the Sum Rule:

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

= - \int \frac{1 0 ^{x}}{1 0 ^{x} + 1} d x + \int 1 d x

\int \frac{1 0 ^{x}}{1 0 ^{x} + 1} d x = \frac{1}{ln ( 10 )} ln ∣ 1 0^{x} + 1 ∣

\int 1 d x = x

= - \frac{1}{ln ( 10 )} ln ∣ 1 0^{x} + 1 ∣ + x

Add a constant to the solution

= - \frac{1}{ln ( 10 )} ln ∣ 1 0^{x} + 1 ∣ + x + C

\frac{d}{d x} (10 cos (30 x - 7))

\int csch^{2} (13 x - \frac{4}{9}) d x

(- 1, - 1), (4, - 6)

6 t^{5}

(1 + t) (8 t^{2} - 1)

What is the integral of 1/(10^x+1) ?
The integral of 1/(10^x+1) is -1/(ln(10))ln|10^x+1|+x+C