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

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

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integral of ln(1+1/x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{x + 1} d x = - ln ∣ x + 1 ∣

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

Simplify

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

Add a constant to the solution

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

x^{2} y^{^{''}} + 17 x y^{^{'}} + 113 y = 0

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

n = 0 \sum \infty \frac{7}{3 ^{n}}

\frac{d}{d x} (cosh (k x))

\int \frac{1}{tan ^{2} ( u ) + 1} d u

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