integral from 0 to x of 1/(0.2x+10)

Lösung

\int_{0}^{x} \frac{1}{0.2 x + 10} d x

5 (ln ∣ 0.2 x + 10 ∣ - 2.30258 \dots)

Schritte zur Lösung

\int_{0}^{x} \frac{1}{0.2 x + 10} d x

Wende U-Substitution an

= \int_{10}^{0.2 x + 10} \frac{1}{0.2 u} d u

Entferne die Konstante:

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

= \frac{1}{0.2} \cdot \int_{10}^{0.2 x + 10} \frac{1}{u} d u

Vereinfache

= 5 \cdot \int_{10}^{0.2 x + 10} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 [ln ∣ u ∣]_{10}^{0.2 x + 10}

Berechne die Grenzen:

ln ∣ 0.2 x + 10 ∣ - 2.30258 \dots

= 5 (ln ∣ 0.2 x + 10 ∣ - 2.30258 \dots)

t an g e n t 1 - x^{2}

n = 1 \sum \infty n^{2} e^{- n}

x \to 0 lim (\frac{cos ( 5 x ) tan ( 5 x )}{3 x})

\int - \frac{1}{( t + 2 ) ^{2}} d t

\frac{\partial}{\partial x} (sin^{2} (a x))