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

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

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integral of 1/(1+1/x)

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

\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x}{( x + 1 ) ^{2}} d x = ln ∣ x + 1 ∣ + \frac{1}{x + 1}

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

Simplify

= \frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}

Add a constant to the solution

= \frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

\frac{d}{d x} (4 x + 2)

\frac{d}{d x} (\frac{8}{3 + x ^{2}})

\int x^{\frac{1}{2}} d x

\frac{d}{d x} (\frac{y - 2 x}{2 y - x})

\frac{1}{1 + 2 x}

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