integral of (1+x^2)2x

Lösung

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

x^{2} + \frac{x ^{4}}{2} + C

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Multipliziere aus

(1 + x^{2}) x : x + x^{3}

= 2 \cdot \int x + x^{3} 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

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

= 2 (\frac{x ^{2}}{2} + \frac{x ^{4}}{4})

Vereinfache

2 (\frac{x ^{2}}{2} + \frac{x ^{4}}{4}) : x^{2} + \frac{x ^{4}}{2}

= x^{2} + \frac{x ^{4}}{2}

Füge eine Konstante zur Lösung hinzu

= x^{2} + \frac{x ^{4}}{2} + C

\int \frac{1}{x ^{2} 16 + 9 x ^{2}} d x

y^{^{''}} - 6 y = 9 t

\int 5 x^{2} + 9 d x

t v^{^{'}} (1 - t^{2}) + 2 v = 0

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