integral of 1/(1+\frac{1){x^2}}

Lösung

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

- arctan (x) + x + C

Schritte zur Lösung

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

Vereinfache

\frac{1}{1 + \frac{1}{x ^{2}}} : \frac{x ^{2}}{x ^{2} + 1}

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

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

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

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

\int 1 d x = x

= - arctan (x) + x

Füge eine Konstante zur Lösung hinzu

= - arctan (x) + x + C

\int \frac{sec ^{2} ( 5 x )}{1 - tan ( 5 x )} d x

\int (1 + tan (x)) sec^{2} (x) d x

\int \frac{sin ^{2} ( x )}{sin ( x )} d x

\int \frac{8 x ^{2}}{e ^{2 x}} d x

\int 2 x 1 + x d x