integral of xe^{2x+1}

Lösung

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

\frac{1}{2} e^{2 x + 1} x - \frac{1}{4} e^{2 x + 1} + C

Schritte zur Lösung

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

Wende die partielle Integration an

= \frac{1}{2} e^{2 x + 1} x - \int \frac{1}{2} e^{2 x + 1} d x

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

= \frac{1}{2} e^{2 x + 1} x - \frac{1}{4} e^{2 x + 1}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} e^{2 x + 1} x - \frac{1}{4} e^{2 x + 1} + C

x \to 0 lim (\frac{x}{x ^{2} - 6 x})

t an g e n t f (x) = \frac{5 x - 7}{x + 1}, at x = 0

t an g e n t f (x) = x^{2} - 2, at x = - 5

x \to 0 lim (2 x^{2} - 9 x + 5)

d er i v a t i v e \frac{1}{( x ^{2} + 1 ) ^{4}}