What is the integral of (x^2+4x)/((x+2)^2) ?

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

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integral of (x^2+4x)/((x+2)^2)

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

x + \frac{4}{x + 2} + 2 + C

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{4 x}{( x + 2 ) ^{2}} d x = 4 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

= x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} + 4 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

Simplify

x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} + 4 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) : x + \frac{4}{x + 2} + 2

= x + \frac{4}{x + 2} + 2

Add a constant to the solution

= x + \frac{4}{x + 2} + 2 + C

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