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

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

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

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

y + 2 ln ∣ y ∣ - \frac{1}{y} + C

Solution steps

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

Expand

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

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

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 y + \int \frac{2}{y} d y + \int \frac{1}{y ^{2}} d y

\int 1 d y = y

\int \frac{2}{y} d y = 2 ln ∣ y ∣

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

= y + 2 ln ∣ y ∣ - \frac{1}{y}

Add a constant to the solution

= y + 2 ln ∣ y ∣ - \frac{1}{y} + C

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