integral of u-1

Lösung

\int u - 1 d u

\frac{u ^{2}}{2} - u + C

Schritte zur Lösung

\int u - 1 d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u - \int 1 d u

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

\int 1 d u = u

= \frac{u ^{2}}{2} - u

Füge eine Konstante zur Lösung hinzu

= \frac{u ^{2}}{2} - u + C

\int \frac{tan ( ln ( 4 x + 3 ) )}{4 x + 3} d x

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{3} z^{2}))

(x + \frac{1}{x})^{^{'}}

d er i v a t i v e [x + (x + sin^{2} (x))^{3}]^{5}

\frac{d}{d x} (\frac{ln ( x + 1 )}{4 x + 2})