integral from 0 to 1/2 of 1/(1+x^4)

Lösung

\int_{0}^{\frac{1}{2}} \frac{1}{1 + x ^{4}} d x

\frac{4 arctan ( \frac{2 + 2}{2} ) + 4 arctan ( \frac{2 - 2}{2} ) + 2 ( - ln ( 2 ) + ln ( \frac{5}{2} + 2 ) ) + 2 ( ln ( 2 ) - ln ( \frac{5}{2} - 2 ) )}{8 2}

Dezimale

0.49395 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{2}} \frac{1}{1 + x ^{4}} d x

Ermittle den Partialbruch von

\frac{1}{1 + x ^{4}} : \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )}

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{\frac{1}{2}} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{2 ( ln ( 2 + \frac{5}{2} ) - ln ( 2 ) ) + 4 arctan ( \frac{1 + 2}{2} ) - π}{8 2}

\int_{0}^{\frac{1}{2}} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{2 ( ln ( 2 ) - ln ( - 2 + \frac{5}{2} ) ) + 4 arctan ( \frac{1 - 2}{2} ) + π}{8 2}

= \frac{2 ( ln ( 2 + \frac{5}{2} ) - ln ( 2 ) ) + 4 arctan ( \frac{1 + 2}{2} ) - π}{8 2} + \frac{2 ( ln ( 2 ) - ln ( - 2 + \frac{5}{2} ) ) + 4 arctan ( \frac{1 - 2}{2} ) + π}{8 2}

Vereinfache

\frac{2 ( ln ( 2 + \frac{5}{2} ) - ln ( 2 ) ) + 4 arctan ( \frac{1 + 2}{2} ) - π}{8 2} + \frac{2 ( ln ( 2 ) - ln ( - 2 + \frac{5}{2} ) ) + 4 arctan ( \frac{1 - 2}{2} ) + π}{8 2} : \frac{4 arctan ( \frac{2 + 2}{2} ) + 4 arctan ( \frac{2 - 2}{2} ) + 2 ( - ln ( 2 ) + ln ( \frac{5}{2} + 2 ) ) + 2 ( ln ( 2 ) - ln ( \frac{5}{2} - 2 ) )}{8 2}

= \frac{4 arctan ( \frac{2 + 2}{2} ) + 4 arctan ( \frac{2 - 2}{2} ) + 2 ( - ln ( 2 ) + ln ( \frac{5}{2} + 2 ) ) + 2 ( ln ( 2 ) - ln ( \frac{5}{2} - 2 ) )}{8 2}

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

\int_{- π}^{π} t d t

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

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