integral from 1 to infinity of 8/(x^2+1)

Lösung

\int_{1}^{\infty} \frac{8}{x ^{2} + 1} d x

2 π

Dezimale

6.28318 \dots

Schritte zur Lösung

\int_{1}^{\infty} \frac{8}{x ^{2} + 1} d x

Berechne das unbestimmte Integral:

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x) + C

Berechne die Grenzen:

\int_{1}^{\infty} \frac{8}{x ^{2} + 1} d x = 4 π - 2 π

= 4 π - 2 π

Vereinfache

= 2 π

\int (csc^{2} (x) - 2 e^{x}) d x

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

d er i v a t i v e f (x) = 2 x^{2} e^{x} (x^{2} - 3 x + 1)

\int 2 x \cdot e^{x} d x

\int_{1}^{4} \frac{x - 7}{x} d x