What is the integral of (((1+x))/x)^2 ?

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

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integral of (((1+x))/x)^2

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

- \frac{1}{x} + 2 ln ∣ x ∣ + x + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} + 2 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{1}{x} + 2 ln ∣ x ∣ + x + C

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