What is the integral from 1 to 2 of (x^2-x-5)/(x+2) ?

The integral from 1 to 2 of (x^2-x-5)/(x+2) is 2ln(2)-ln(3)-3/2

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integral from 1 to 2 of (x^2-x-5)/(x+2)

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

2 ln (2) - ln (3) - \frac{3}{2}

Decimal

- 1.21231 \dots

Solution steps

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

Expand

\frac{x ^{2} - x - 5}{x + 2} : \frac{x ^{2}}{x + 2} - \frac{x}{x + 2} - \frac{5}{x + 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_{1}^{2} \frac{x ^{2}}{x + 2} d x - \int_{1}^{2} \frac{x}{x + 2} d x - \int_{1}^{2} \frac{5}{x + 2} d x

\int_{1}^{2} \frac{x ^{2}}{x + 2} d x = 4 (2 ln (2) - ln (3)) - \frac{1}{2}

\int_{1}^{2} \frac{x}{x + 2} d x = 1 - 2 (2 ln (2) - ln (3))

\int_{1}^{2} \frac{5}{x + 2} d x = 5 (2 ln (2) - ln (3))

= 4 (2 ln (2) - ln (3)) - \frac{1}{2} - (1 - 2 (2 ln (2) - ln (3))) - 5 (2 ln (2) - ln (3))

Simplify

4 (2 ln (2) - ln (3)) - \frac{1}{2} - (1 - 2 (2 ln (2) - ln (3))) - 5 (2 ln (2) - ln (3)) : 2 ln (2) - ln (3) - \frac{3}{2}

= 2 ln (2) - ln (3) - \frac{3}{2}

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 3

\int x ln (13 x) d x

\frac{9}{22 ( 3 s - 2 )}

\int \frac{3 ^{x}}{1 + 3 ^{x}} d x

f (x) = 3 x^{2} + 5 x, at x = - 2

What is the integral from 1 to 2 of (x^2-x-5)/(x+2) ?
The integral from 1 to 2 of (x^2-x-5)/(x+2) is 2ln(2)-ln(3)-3/2