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

The integral of (10^x+1)/(10^x-1) is (2ln|10^x-1|-ln(10)x)/(ln(10))+C

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integral of (10^x+1)/(10^x-1)

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

\frac{2 ln ∣ 1 0 ^{x} - 1 ∣ - ln ( 10 ) x}{ln ( 10 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{ln ( 10 ) u ( u - 1 )} d u

= \int \frac{u + 1}{ln ( 10 ) u ( u - 1 )} d u

Take the constant out:

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

= \frac{1}{ln ( 10 )} \cdot \int \frac{u + 1}{u ( u - 1 )} d u

Expand

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

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 ( 10 )} (\int \frac{u}{u ( u - 1 )} d u + \int \frac{1}{u ( u - 1 )} d u)

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

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

= \frac{1}{ln ( 10 )} (ln ∣ u - 1 ∣ - ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = 1 0^{x}

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

Simplify

\frac{1}{ln ( 10 )} (ln ∣ 1 0^{x} - 1 ∣ - ln ∣ 1 0^{x} ∣ + ln ∣ 1 0^{x} - 1 ∣) : \frac{2 ln ∣ 1 0 ^{x} - 1 ∣ - ln ( 10 ) x}{ln ( 10 )}

= \frac{2 ln ∣ 1 0 ^{x} - 1 ∣ - ln ( 10 ) x}{ln ( 10 )}

Add a constant to the solution

= \frac{2 ln ∣ 1 0 ^{x} - 1 ∣ - ln ( 10 ) x}{ln ( 10 )} + C

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What is the integral of (10^x+1)/(10^x-1) ?
The integral of (10^x+1)/(10^x-1) is (2ln|10^x-1|-ln(10)x)/(ln(10))+C