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

The integral of ((2y+1)^2)/y is 2y^2+4y+ln|y|+1+C

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

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

2 y^{2} + 4 y + ln ∣ y ∣ + 1 + C

Solution steps

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

Apply Integration By Parts

: (2 y + 1)^{2} - \int \frac{4 y ^{2} - 1}{y} d y

= (2 y + 1)^{2} - \int \frac{4 y ^{2} - 1}{y} d y

\int \frac{4 y ^{2} - 1}{y} d y = 2 y^{2} - ln ∣ y ∣

= (2 y + 1)^{2} - (2 y^{2} - ln ∣ y ∣)

Simplify

(2 y + 1)^{2} - (2 y^{2} - ln ∣ y ∣) : 2 y^{2} + 4 y + ln ∣ y ∣ + 1

= 2 y^{2} + 4 y + ln ∣ y ∣ + 1

Add a constant to the solution

= 2 y^{2} + 4 y + ln ∣ y ∣ + 1 + C

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What is the integral of ((2y+1)^2)/y ?
The integral of ((2y+1)^2)/y is 2y^2+4y+ln|y|+1+C