integral of pi/4 x-xarctan(x)

Lösung

\int \frac{π}{4} x - x arctan (x) d x

\frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x + C

Schritte zur Lösung

\int \frac{π}{4} x - x arctan (x) 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 \frac{π}{4} x d x - \int x arctan (x) d x

\int \frac{π}{4} x d x = \frac{π x ^{2}}{8}

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

= \frac{π x ^{2}}{8} - (\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Vereinfache

\frac{π x ^{2}}{8} - (\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x

= \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x

Füge eine Konstante zur Lösung hinzu

= \frac{π x ^{2}}{8} - \frac{1}{2} x^{2} arctan (x) - \frac{1}{2} arctan (x) + \frac{1}{2} x + C

\int cos (8 x^{2} + π) d x

\int (\frac{x ^{6} + 3 x ^{4} - 2 x}{x ^{4}}) d x

2 x (1 + y^{2}) d x - y (1 + 2 x^{2}) d y = 0

\int \frac{1}{x ^{4} + 1} d x

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