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

integral

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

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

\frac{sec ^{2} ( x )}{tan ( x )}

\frac{sin ^{2} ( x )}{cos ( x )}

\frac{( 2 + e ^{x} )}{e ^{x}}

(16 - x^{2})^{\frac{1}{2}}

\frac{a}{( a ^{2} + x ^{2} )}

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