integral from 1 to 2e of ln(2x+1)

Lösung

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

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

Dezimale

8.60422 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{3}^{4 e + 1} ln (u) \frac{1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{3}^{4 e + 1} ln (u) d u

Wende die partielle Integration an

= \frac{1}{2} [u ln (u) - \int 1 d u]_{3}^{4 e + 1}

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{3}^{4 e + 1}

Berechne die Grenzen:

4 e ln (4 e + 1) + ln (4 e + 1) + 2 - 4 e - 3 ln (3)

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

Vereinfache

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

\int_{3}^{7} (3 - x) (x - 7) d x

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

\int_{- 1}^{3} 2 x + 6 d x

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

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