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

Lösung

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

π - \frac{1}{2} ln (2)

Dezimale

2.79501 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{4 - x}{1 + x ^{2}} : \frac{4}{1 + x ^{2}} - \frac{x}{1 + x ^{2}}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{1} \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (2)

= π - \frac{1}{2} ln (2)

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) + 5} d x

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

\frac{\partial}{\partial x} (x cos (\frac{x}{y}))

x \to 0 lim (x^{3} - 2 x^{3} - 4 x + 7)

\frac{\partial}{\partial x} (e^{- 3 x^{2} - y^{2}})