integral of t/(1-t^8)

Lösung

\int \frac{t}{1 - t ^{8}} d t

\frac{1}{2} (\frac{1}{2} arctan (t^{2}) + \frac{1}{4} ln t^{2} + 1 - \frac{1}{4} ln t^{2} - 1) + C

Schritte zur Lösung

\int \frac{t}{1 - t ^{8}} d t

Wende U-Substitution an

= \int \frac{1}{2 ( 1 - u ^{4} )} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{1 - u ^{4}} d u

Ermittle den Partialbruch von

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

= \frac{1}{2} \cdot \int \frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )} d u

Wende die Summenregel an:

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

= \frac{1}{2} (\int \frac{1}{2 ( u ^{2} + 1 )} d u + \int \frac{1}{4 ( u + 1 )} d u - \int \frac{1}{4 ( u - 1 )} d u)

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

= \frac{1}{2} (\frac{1}{2} arctan (u) + \frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u - 1 ∣)

Setze in

u = t^{2}

ein

= \frac{1}{2} (\frac{1}{2} arctan (t^{2}) + \frac{1}{4} ln t^{2} + 1 - \frac{1}{4} ln t^{2} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (\frac{1}{2} arctan (t^{2}) + \frac{1}{4} ln t^{2} + 1 - \frac{1}{4} ln t^{2} - 1) + C

\frac{\partial}{\partial x} ((x - y)^{10})

a re a x^{4}, x^{2}

d er i v a t i v e f (x) = d e^{- c x}

\int \frac{1 - x ^{2}}{2 x} d x

f (t) = e^{3 t}