integral of 1+3x

Lösung

\int 1 + 3 x d x

x + \frac{3 x ^{2}}{2} + C

Schritte zur Lösung

\int 1 + 3 x 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 1 d x + \int 3 x d x

\int 1 d x = x

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

= x + \frac{3 x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= x + \frac{3 x ^{2}}{2} + C

x (\frac{d y}{d x}) = y ln (x y)

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 500 x)

\frac{d}{d t} (b (t) ln (t) + c t^{2})

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

\frac{d}{d x} (\frac{e ^{x} \cdot cos ( x )}{e ^{x} - 1})