integral of (2x+1)^2

Lösung

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

\frac{4 x ^{3}}{3} + 2 x^{2} + x + C

Schritte zur Lösung

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

Multipliziere aus

(2 x + 1)^{2} : 4 x^{2} + 4 x + 1

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 1 d x = x

= \frac{4 x ^{3}}{3} + 2 x^{2} + x

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{3}}{3} + 2 x^{2} + x + C

\frac{d}{d x} (e^{- x} cos (y) + e^{- y} cos (x))

\int (x^{2} - 3)^{5} (2 x) d x

\int \frac{cot ( x )}{cos ( x )} d x

(1 + x) y^{^{'}} = y + x

d er i v a t i v e y = 3 arctan (x + 1 + x^{2})