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

Lösung

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

\frac{1}{2} arctan (16)

Dezimale

0.75418 \dots

Schritte zur Lösung

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

Berechne das unbestimmte Integral:

\int \frac{x}{1 + x ^{4}} d x = \frac{1}{2} arctan (x^{2}) + C

Berechne die Grenzen:

\int_{0}^{4} \frac{x}{1 + x ^{4}} d x = \frac{1}{2} arctan (16) - 0

= \frac{1}{2} arctan (16) - 0

Vereinfache

= \frac{1}{2} arctan (16)

\int_{- 4}^{- 2} - x d x

\int_{2}^{4} (1 - e^{- x}) d x

\int_{0}^{3} 3 x (1 + 3 x^{2})^{2} d x

\int_{y}^{\infty} x e^{- x} d x

\int_{1}^{2} 6 x^{2} - 3 x d x