integral from 3 to u of 1/(x+x^2)

Lösung

\int_{3}^{u} \frac{1}{x + x ^{2}} d x

ln ∣ u ∣ - ln (3) - ln ∣ u + 1 ∣ + 2 ln (2)

Schritte zur Lösung

\int_{3}^{u} \frac{1}{x + x ^{2}} d x

Ermittle den Partialbruch von

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

= \int_{3}^{u} \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_{3}^{u} \frac{1}{x} d x - \int_{3}^{u} \frac{1}{x + 1} d x

\int_{3}^{u} \frac{1}{x} d x = ln ∣ u ∣ - ln (3)

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

= ln ∣ u ∣ - ln (3) - (ln ∣ u + 1 ∣ - 2 ln (2))

Vereinfache

= ln ∣ u ∣ - ln (3) - ln ∣ u + 1 ∣ + 2 ln (2)

\int_{1}^{9} \frac{x - 6}{x} d x

\int_{- π}^{π} x sin (k x) d x

\int_{0}^{5} x (3) d x

\int_{- 1}^{3} (x^{3} - 1) d x

\int_{0}^{2 π} 6 cos^{2} (x) d x