integral from 0 to pi/4 of arctan(x)

Lösung

\int_{0}^{\frac{π}{4}} arctan (x) d x

\frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

Dezimale

0.28265 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} arctan (x) d x

Wende die partielle Integration an

= [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{0}^{\frac{π}{4}}

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{\frac{π}{4}}

Vereinfache

[x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{\frac{π}{4}} : [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{\frac{π}{4}}

Berechne die Grenzen:

\frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

= \frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

\int_{3}^{6} x^{2} - 6 x + 9 d x

\int_{0}^{2 x} e^{t^{2}} d t

\int_{\frac{π}{6}}^{\frac{π}{4}} sin (x) d x

\int_{1}^{2} x e^{- x} d x

\int_{8}^{17} x x^{2} - 64 d x