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

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

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

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

2 x^{2} + 4 x + ln ∣ x ∣ + 1 + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= 2 x^{2} + 4 x + ln ∣ x ∣ + 1

Add a constant to the solution

= 2 x^{2} + 4 x + ln ∣ x ∣ + 1 + C

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