integral from 2 to 3 of 6/(2x^2+2x-4)

Lösung

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

3 ln (2) - ln (5)

Dezimale

0.47000 \dots

Schritte zur Lösung

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{2}^{3} \frac{1}{2 x ^{2} + 2 x - 4} d x

Ermittle den Partialbruch von

\frac{1}{2 x ^{2} + 2 x - 4} : \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 2 )}

= 6 \cdot \int_{2}^{3} \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 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

= 6 (\int_{2}^{3} \frac{1}{6 ( x - 1 )} d x - \int_{2}^{3} \frac{1}{6 ( x + 2 )} d x)

\int_{2}^{3} \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln (2)

\int_{2}^{3} \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} (ln (5) - 2 ln (2))

= 6 (\frac{1}{6} ln (2) - \frac{1}{6} (ln (5) - 2 ln (2)))

Vereinfache

6 (\frac{1}{6} ln (2) - \frac{1}{6} (ln (5) - 2 ln (2))) : 3 ln (2) - ln (5)

= 3 ln (2) - ln (5)

\int_{0}^{1} 3 x^{2} x^{3} - 5 d x

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

\int_{0}^{2} 5 x^{2} - 11 x d x

\int_{1}^{e} (22 + ln (x))^{2} d x

\int_{0}^{1000} e^{4 z} - 5 d z