integral from 0 to 1 of ((4t^3)/(1+t^4))

Lösung

\int_{0}^{1} (\frac{4 t ^{3}}{1 + t ^{4}}) d t

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{4 t ^{3}}{1 + t ^{4}} d t

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} \frac{t ^{3}}{1 + t ^{4}} d t

Wende U-Substitution an

= 4 \cdot \int_{1}^{2} \frac{1}{4 u} d u

Entferne die Konstante:

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

= 4 \cdot \frac{1}{4} \cdot \int_{1}^{2} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{2}

Vereinfache

4 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Berechne die Grenzen:

ln (2)

= ln (2)

\int \frac{( x ^{2} )}{( 3 + 4 x - 4 x ^{2} ) ^{\frac{3}{2}}} d x

s l o p e (0, - 3), (2, 0)

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

\frac{d y}{d x} = (\frac{2 y + 5}{4 x + 9})^{2}

\int_{0}^{\infty} \int_{0}^{\infty} e^{- x - y} d x d y