integral from 0 to 4 of (x+1)/(x^2+2x+1)

Lösung

\int_{0}^{4} \frac{x + 1}{x ^{2} + 2 x + 1} d x

ln (5)

Dezimale

1.60943 \dots

Schritte zur Lösung

\int_{0}^{4} \frac{x + 1}{x ^{2} + 2 x + 1} d x

Wende U-Substitution an

= \int_{1}^{25} \frac{1}{2 u} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{25} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{25}

Berechne die Grenzen:

2 ln (5)

= \frac{1}{2} \cdot 2 ln (5)

Vereinfache

= ln (5)

\int_{0}^{0.5} e^{- x^{3}} d x

\int_{- 3}^{4} ∣ x + 1 ∣ d x

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

\int_{2}^{3} \frac{7}{x} + x^{- 1} d x

\int_{- 1}^{0} (2 x + 1)^{4} (x + 1) d x