integral from 1 to 2 of (x^3+1)/(x^4+4x)

Lösung

\int_{1}^{2} \frac{x ^{3} + 1}{x ^{4} + 4 x} d x

\frac{1}{4} (ln (24) - ln (5))

Dezimale

0.39215 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{x ^{3} + 1}{x ^{4} + 4 x} d x

Wende U-Substitution an

= \int_{5}^{24} \frac{1}{4 u} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{5}^{24} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{5}^{24}

Berechne die Grenzen:

ln (24) - ln (5)

= \frac{1}{4} (ln (24) - ln (5))

\int_{0}^{y^{2}} x^{2} y d x

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

\int_{- 3}^{3} e^{- x^{2}} d x

\int_{0}^{\frac{π}{2}} (\frac{1}{2} + cos (x)) d x

\int_{1}^{4} (x^{3} + 2 x + 5) d x