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

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

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

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

x + 2 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Expand

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

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= x + 2 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= x + 2 ln ∣ x ∣ - \frac{1}{x} + C

f (x) = ln ((20 + 2 x)^{4})

y^{^{'}} = 0.05 y - 48000

\int \frac{cos ( \frac{π}{x ^{12}} )}{x ^{13}} d x

4 (x + 2)^{3}

\int \frac{a}{s ^{2} - a ^{2}} d s

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