integral of (x^4+3x^2+3)/(x^2+1)

Lösung

\int \frac{x ^{4} + 3 x ^{2} + 3}{x ^{2} + 1} d x

\frac{x ^{3}}{3} + 2 x + arctan (x) + C

Schritte zur Lösung

\int \frac{x ^{4} + 3 x ^{2} + 3}{x ^{2} + 1} d x

Multipliziere aus

\frac{x ^{4} + 3 x ^{2} + 3}{x ^{2} + 1} : \frac{x ^{4}}{x ^{2} + 1} + \frac{3 x ^{2}}{x ^{2} + 1} + \frac{3}{x ^{2} + 1}

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 ^{4}}{x ^{2} + 1} d x + \int \frac{3 x ^{2}}{x ^{2} + 1} d x + \int \frac{3}{x ^{2} + 1} d x

\int \frac{x ^{4}}{x ^{2} + 1} d x = \frac{x ^{3}}{3} - x + arctan (x)

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

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

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

Vereinfache

\frac{x ^{3}}{3} - x + arctan (x) + 3 (- arctan (x) + x) + 3 arctan (x) : \frac{x ^{3}}{3} + 2 x + arctan (x)

= \frac{x ^{3}}{3} + 2 x + arctan (x)

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} + 2 x + arctan (x) + C

\int (x^{2} - 4 x)^{5} \cdot (4 x - 8) d x

\int π^{2} cos (n x) d x

\int (1 - x)^{n - 1} d x

\int \frac{3 y + 1}{y} d y

\int \frac{3}{4 + x} d x