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

Lösung

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

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

Dezimale

- 1.21231 \dots

Schritte zur Lösung

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

Multipliziere aus

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

Wende die Summenregel an:

\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))

Vereinfache

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}

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

\int x ln (13 x) d x

in v erse l a pl a ce \frac{9}{22 ( 3 s - 2 )}

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

t an g e n t f (x) = 3 x^{2} + 5 x, at x = - 2