integral from 2 to 1/x of arctan(t)

Lösung

\int_{2}^{\frac{1}{x}} arctan (t) d t

\frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

Schritte zur Lösung

\int_{2}^{\frac{1}{x}} arctan (t) d t

Wende die partielle Integration an

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

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

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

Berechne die Grenzen:

\frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

= \frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

Vereinfache

= \frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

\int_{- 5}^{0} (4 + 25 - x^{2}) d x

x \to 8 lim (\frac{x + 9}{x - 8})

y^{^{''}} + 256 y = 7 cos (5 t)

in v erse l a pl a ce \frac{10}{s ^{2} + 25} + \frac{4}{s - 3}

\int \frac{x ^{2} - 25}{x} d x