integral of x^3-3x+\sqrt[4]{x}-5

Lösung

\int x^{3} - 3 x + 4 x - 5 d x

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + \frac{4}{5} x^{\frac{5}{4}} - 5 x + C

Schritte zur Lösung

\int x^{3} - 3 x + 4 x - 5 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^{3} d x - \int 3 x d x + \int 4 x d x - \int 5 d x

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

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

\int 4 x d x = \frac{4}{5} x^{\frac{5}{4}}

\int 5 d x = 5 x

= \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + \frac{4}{5} x^{\frac{5}{4}} - 5 x

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + \frac{4}{5} x^{\frac{5}{4}} - 5 x + C

n = 0 \sum \infty 3^{n} 8^{- n + 1}

f (x) = \frac{5}{x}

\int x^{- \frac{3}{4}} d x

\int_{0}^{π} sin^{2} (3 x) d x

y^{^{''}} + 121 y = sec (11 x), y (0) = - 5, y^{^{'}} (0) = 6