integral from 1 to infinity of 1/(x^2+5)

Lösung

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x

\frac{π - 2 arctan ( \frac{5}{5} )}{2 5}

Dezimale

0.51441 \dots

Schritte zur Lösung

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x

Berechne das unbestimmte Integral:

\int \frac{1}{x ^{2} + 5} d x = \frac{1}{5} arctan (\frac{x}{5}) + C

Berechne die Grenzen:

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x = \frac{π}{2 5} - \frac{1}{5} arctan (\frac{1}{5})

= \frac{π}{2 5} - \frac{1}{5} arctan (\frac{1}{5})

Vereinfache

= \frac{π - 2 arctan ( \frac{5}{5} )}{2 5}

l a pl a ce t r an s f or m 5^{t}

t an g e n t f (x) = x + 3, at x = 1

\frac{d}{d x} (\frac{ln ( x ^{5} )}{ln ( x ^{4} )})

\int_{- 2}^{2} (8 - 2 y^{2}) d y

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