What is the integral of (x^2-3x-10)/(x+2) ?

The integral of (x^2-3x-10)/(x+2) is (x^2)/2-5x+C

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integral of (x^2-3x-10)/(x+2)

\int \frac{x ^{2} - 3 x - 10}{x + 2} d x

\frac{x ^{2}}{2} - 5 x + C

Solution steps

\int \frac{x ^{2} - 3 x - 10}{x + 2} d x

Simplify

\frac{x ^{2} - 3 x - 10}{x + 2} : x - 5

= \int x - 5 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 x d x - \int 5 d x

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

\int 5 d x = 5 x

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

Add a constant to the solution

= \frac{x ^{2}}{2} - 5 x + C

x \to 0 + lim (x cos (\frac{3}{x}))

x \to 1 + lim (2 x - 1)

\frac{d}{d x} (3 x ln (x))

n = 1 \sum \infty 2 - 8 (0.9)^{n}

\int x^{2} x^{3} + 17 d x

What is the integral of (x^2-3x-10)/(x+2) ?
The integral of (x^2-3x-10)/(x+2) is (x^2)/2-5x+C