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

Lösung

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

2 ln (2) - ln (3)

Dezimale

0.28768 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{x ^{2} + x} : \frac{1}{x} - \frac{1}{x + 1}

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

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

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

= ln (2) - (ln (3) - ln (2))

Vereinfache

= 2 ln (2) - ln (3)

\int_{0}^{3} (32 x^{3} + 4 x) d x

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

\int_{2}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

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

\int_{0}^{l n (x)} e^{x} - 1 d x