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

Lösung

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

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{2}

Berechne die Grenzen:

ln (2)

= ln (2)

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

\frac{\partial}{\partial y} (16 - x^{2} - 4 y)

y^{^{''}} - y = sin (2 t)

d er i v a t i v e f (x) = \frac{e ^{x}}{x + 3}

in v erse l a pl a ce \frac{( s + 5 )}{( s ^{2} + 4 s + 13 )}