integral of (x-arctan(2x))/(1+4x^2)

Lösung

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

\frac{1}{8} ln 1 + 4 x^{2} - \frac{1}{4} arctan^{2} (2 x) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{x}{1 + 4 x ^{2}} d x = \frac{1}{8} ln 1 + 4 x^{2}

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

= \frac{1}{8} ln 1 + 4 x^{2} - \frac{1}{4} arctan^{2} (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln 1 + 4 x^{2} - \frac{1}{4} arctan^{2} (2 x) + C

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