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

The integral of ((2x+1)^2)/(x+2) is 2x^2-4x+9ln|x+2|+C

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

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

2 x^{2} - 4 x + 9 ln ∣ x + 2 ∣ + C

Solution steps

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

Long division

\frac{( 2 x + 1 ) ^{2}}{x + 2} : 4 x - 4 + \frac{9}{x + 2}

= \int 4 x - 4 + \frac{9}{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 4 x d x - \int 4 d x + \int \frac{9}{x + 2} d x

\int 4 x d x = 2 x^{2}

\int 4 d x = 4 x

\int \frac{9}{x + 2} d x = 9 ln ∣ x + 2 ∣

= 2 x^{2} - 4 x + 9 ln ∣ x + 2 ∣

Add a constant to the solution

= 2 x^{2} - 4 x + 9 ln ∣ x + 2 ∣ + C

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What is the integral of ((2x+1)^2)/(x+2) ?
The integral of ((2x+1)^2)/(x+2) is 2x^2-4x+9ln|x+2|+C