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

Lösung

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

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

Schritte zur Lösung

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

Lange Division

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

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

\int 1 d x = x

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

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

Füge eine Konstante zur Lösung hinzu

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

\int - 6 x^{9} d x

\int [ln (x + 1 + x^{2})]^{2} d x

\int \frac{2 x ^{2}}{x ^{3} - 1} d x

\int x 100 - x^{2} d x

\int (7 x^{3} - 4 x^{2} + 11 x - e^{2} + 17) d x