What is the integral of-(x^2+11x+28)/((x+2)^2) ?

The integral of-(x^2+11x+28)/((x+2)^2) is -x-7ln|x+2|+(10)/(x+2)-2+C

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integral of-(x^2+11x+28)/((x+2)^2)

\int - \frac{x ^{2} + 11 x + 28}{( x + 2 ) ^{2}} d x

- x - 7 ln ∣ x + 2 ∣ + \frac{10}{x + 2} - 2 + C

Solution steps

\int - \frac{x ^{2} + 11 x + 28}{( x + 2 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{x ^{2} + 11 x + 28}{( x + 2 ) ^{2}} d x

Expand

\frac{x ^{2} + 11 x + 28}{( x + 2 ) ^{2}} : \frac{x ^{2}}{( x + 2 ) ^{2}} + \frac{11 x}{( x + 2 ) ^{2}} + \frac{28}{( 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{11 x}{( x + 2 ) ^{2}} d x + \int \frac{28}{( 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{11 x}{( x + 2 ) ^{2}} d x = 11 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

\int \frac{28}{( x + 2 ) ^{2}} d x = - \frac{28}{x + 2}

= - (x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} + 11 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{28}{x + 2})

Simplify

- (x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} + 11 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{28}{x + 2}) : - x - 7 ln ∣ x + 2 ∣ + \frac{10}{x + 2} - 2

= - x - 7 ln ∣ x + 2 ∣ + \frac{10}{x + 2} - 2

Add a constant to the solution

= - x - 7 ln ∣ x + 2 ∣ + \frac{10}{x + 2} - 2 + C

\frac{d}{d x} (\frac{1}{x ^{2} + 5})

f (x) = - 16 t^{2}

\frac{s - 1}{2 ( s ^{2} + 1 )}

2 x^{2} - 5 x + 3

sin (x - x^{2})

What is the integral of-(x^2+11x+28)/((x+2)^2) ?
The integral of-(x^2+11x+28)/((x+2)^2) is -x-7ln|x+2|+(10)/(x+2)-2+C