integral of (x^4+x^2+1)/(x^2+1)

Lösung

\int \frac{x ^{4} + x ^{2} + 1}{x ^{2} + 1} d x

\frac{x ^{3}}{3} + arctan (x) + C

Schritte zur Lösung

\int \frac{x ^{4} + x ^{2} + 1}{x ^{2} + 1} d x

Lange Division

\frac{x ^{4} + x ^{2} + 1}{x ^{2} + 1} : x^{2} + \frac{1}{x ^{2} + 1}

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

= \frac{x ^{3}}{3} + arctan (x)

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} + arctan (x) + C

\int (2 x^{3} - x)^{4} (6 x^{2} - 1) d x

\int \frac{1}{( v ^{2} - 3 ) ^{\frac{3}{2}}} d v

\int \frac{x + 4}{x ^{3} + 6 x ^{2} + 9 x} d x

\int x I n^{2} x d x

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