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

Lösung

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

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

Schritte zur Lösung

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

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

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

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{x \cdot 1 + x ^{2}} d x

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\int cos (\frac{π}{6} x) d x