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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d y} (- \frac{y}{( x + 2 ) ^{2}})

\frac{d}{d x} (ln (3 x) - e^{s e c (x)})

f (x) = \frac{x + 1}{x ^{2} - 5}

f (t) = e^{t} cos (t)

ma c l a u r in sin (3 x + 2)