integral from 3 to x of 1/(1+t^4)

Lösung

\int_{3}^{x} \frac{1}{1 + t ^{4}} d t

\frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (62 + 20)) + \frac{1}{2} (arctan (2 x + 1) - arctan (32 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (20 - 62)) + \frac{1}{2} (arctan (2 x - 1) - arctan (32 - 1)))

Schritte zur Lösung

\int_{3}^{x} \frac{1}{1 + t ^{4}} d t

Ermittle den Partialbruch von

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

= \int_{3}^{x} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} + \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t

Wende die Summenregel an:

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

= \int_{3}^{x} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t + \int_{3}^{x} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t

\int_{3}^{x} \frac{t + 2}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (62 + 20)) + \frac{1}{2} (arctan (2 x + 1) - arctan (32 + 1)))

\int_{3}^{x} \frac{- t + 2}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (20 - 62)) + \frac{1}{2} (arctan (2 x - 1) - arctan (32 - 1)))

= \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (62 + 20)) + \frac{1}{2} (arctan (2 x + 1) - arctan (32 + 1))) + \frac{1}{2} (- \frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (20 - 62)) + \frac{1}{2} (arctan (2 x - 1) - arctan (32 - 1)))

\int_{3}^{8} x + 1 d x

\int_{- \frac{3 π}{4}}^{0} sin (x) d x

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\int_{0}^{4} y 9 + y^{2} d y