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

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

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

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

\frac{y ^{2}}{2} + 2 y + ln ∣ y ∣ + C

Solution steps

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

Simplify

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

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

Expand

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

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

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

\int 2 d y = 2 y

\int \frac{1}{y} d y = ln ∣ y ∣

= \frac{y ^{2}}{2} + 2 y + ln ∣ y ∣

Add a constant to the solution

= \frac{y ^{2}}{2} + 2 y + ln ∣ y ∣ + C

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