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

Lösung

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

- arctan (x) + x + C

Schritte zur Lösung

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

Vereinfache

x^{2} \frac{1}{x ^{2} + 1} : \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{1}{( 7 x - 6 ) ^{2}} d x

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{9} d x

\int \frac{1}{x ^{2} - 3 x - 18} d x

\int 1 + (\frac{3}{2} x^{\frac{1}{2}})^{2} d x

\int 3 x + 4 x d x