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

Lösung

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

\frac{1}{2} ln R^{2} + 1

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{1 + R^{2}}

Berechne die Grenzen:

ln R^{2} + 1

= \frac{1}{2} ln R^{2} + 1

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

\int \frac{i n ^{2} x}{x} d x

\frac{\partial}{\partial x} (3 x^{4} y^{2} + 1)

x \to 2 lim (\frac{2}{( x - 2 ) ^{2}})

\frac{\partial}{\partial y} (x^{5} + x^{2} y + x + 1)