What is the integral of (x^2-8)/(x+3) ?

The integral of (x^2-8)/(x+3) is 1/2 (x+3)^2-6x-18+ln|x+3|+C

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integral of (x^2-8)/(x+3)

\int \frac{x ^{2} - 8}{x + 3} d x

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

Solution steps

\int \frac{x ^{2} - 8}{x + 3} d x

Expand

\frac{x ^{2} - 8}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{8}{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{8}{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{8}{x + 3} d x = 8 ln ∣ x + 3 ∣

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 8 ln ∣ x + 3 ∣

Simplify

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 8 ln ∣ x + 3 ∣ : \frac{1}{2} (x + 3)^{2} - 6 x - 18 + ln ∣ x + 3 ∣

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

Add a constant to the solution

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

\frac{x ^{2} - 25}{x - 5}

\int (5 e^{2 x}) d x

y^{^{'}} = \frac{y ^{2}}{x}

\int x \cdot 1 - x d x

x y^{^{'}} - y = x^{2} - x + 2

What is the integral of (x^2-8)/(x+3) ?
The integral of (x^2-8)/(x+3) is 1/2 (x+3)^2-6x-18+ln|x+3|+C