integral of sqrt(1+x^2)*x^5

Lösung

\int 1 + x^{2} \cdot x^{5} d x

\frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

Schritte zur Lösung

\int 1 + x^{2} x^{5} d x

Wende U-Substitution an

= \int \frac{u ^{2} u + 1}{2} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{2} u + 1 d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int (v - 1)^{2} v d v

Multipliziere aus

(v - 1)^{2} v : v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + v

= \frac{1}{2} \cdot \int v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + v d v

Wende die Summenregel an:

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

= \frac{1}{2} (\int v^{\frac{5}{2}} d v - \int 2 v^{\frac{3}{2}} d v + \int v d v)

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

\int 2 v^{\frac{3}{2}} d v = \frac{4}{5} v^{\frac{5}{2}}

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

= \frac{1}{2} (\frac{2}{7} v^{\frac{7}{2}} - \frac{4}{5} v^{\frac{5}{2}} + \frac{2}{3} v^{\frac{3}{2}})

Ersetze zurück

= \frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}})

Vereinfache

\frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

t an g e n t y = - x^{3} + 2 x^{2} - 4, at x = - 1

d er i v a t i v e - sec^{2} (x)

\frac{d}{d n} (e^{niπ})

y^{^{''}} + y^{^{'}} + y = 0

\int (y^{5} + 5 y) d y