integral 0^{13}x+1(x+1)^2

Lösung

integral

0^{13} x + 1 (x + 1)^{2}

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

Schritte zur Lösung

\int 0^{13} x + 1 \cdot (x + 1)^{2} d x

Vereinfache

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

Multipliziere aus

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

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

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

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{e ^{x}}{( x + 2 )}

in t e g r a l e^{x} cos^{2} (x)

in t e g r a l y = cos (x^{2}) (0.1)

in t e g r a l t^{2} (e^{t})^{2}

in t e g r a l 4 x^{10}