integral from 2 to e of (x^2+3)/(x^3+2x)

Lösung

\int_{2}^{e} \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

\frac{3}{2} (1 - ln (2)) - \frac{1}{4} (ln (e^{2} + 2) - ln (6))

Dezimale

0.34833 \dots

Schritte zur Lösung

\int_{2}^{e} \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

Ermittle den Partialbruch von

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

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{2}^{e} \frac{x}{2 ( x ^{2} + 2 )} d x = \frac{1}{4} (ln (e^{2} + 2) - ln (6))

= \frac{3}{2} (1 - ln (2)) - \frac{1}{4} (ln (e^{2} + 2) - ln (6))

\int_{0}^{1} ln (1 + x^{2}) x d x

\int_{0}^{1} 1 - t d t

\int_{0}^{3} (9 - x^{2}) d x

\int_{3}^{4} \frac{2}{x ^{2}} d x

\int_{0}^{1} (5 x^{2} - 6 x + 6) d x