What is the integral of (x^2-2)/(x+3) ?

The integral of (x^2-2)/(x+3) is 1/2 (x+3)^2-6x-18+7ln|x+3|+C

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integral of (x^2-2)/(x+3)

\int \frac{x ^{2} - 2}{x + 3} d x

\frac{1}{2} (x + 3)^{2} - 6 x - 18 + 7 ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x ^{2} - 2}{x + 3} d x

Expand

\frac{x ^{2} - 2}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{2}{x + 3}

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 + 3} d x - \int \frac{2}{x + 3} d x

\int \frac{x ^{2}}{x + 3} d x = \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣

\int \frac{2}{x + 3} d x = 2 ln ∣ x + 3 ∣

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 2 ln ∣ x + 3 ∣

Simplify

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 2 ln ∣ x + 3 ∣ : \frac{1}{2} (x + 3)^{2} - 6 x - 18 + 7 ln ∣ x + 3 ∣

= \frac{1}{2} (x + 3)^{2} - 6 x - 18 + 7 ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{2} (x + 3)^{2} - 6 x - 18 + 7 ln ∣ x + 3 ∣ + C

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

\frac{d}{d x} (\frac{( x - 3 ) ( 3 x - 25 )}{x})

f (x) = x^{2} + 5 x - 2, at x = - 1

x \to 0 + lim (\frac{6}{x ^{2} - x})

f (x) = x - 5

What is the integral of (x^2-2)/(x+3) ?
The integral of (x^2-2)/(x+3) is 1/2 (x+3)^2-6x-18+7ln|x+3|+C