integral from 0 to x of v^2arcsin(v^3)

Lösung

\int_{0}^{x} v^{2} arcsin (v^{3}) d v

\frac{x ^{3} arcsin ( x ^{3} ) + 1 - x ^{6} - 1}{3}

Schritte zur Lösung

\int_{0}^{x} v^{2} arcsin (v^{3}) d v

Wende die partielle Integration an

= [\frac{1}{3} v^{3} arcsin (v^{3}) - \int \frac{v ^{5}}{1 - v ^{6}} d v]_{0}^{x}

\int \frac{v ^{5}}{1 - v ^{6}} d v = - \frac{1}{3} 1 - v^{6}

= [\frac{1}{3} v^{3} arcsin (v^{3}) - (- \frac{1}{3} 1 - v^{6})]_{0}^{x}

Vereinfache

[\frac{1}{3} v^{3} arcsin (v^{3}) - (- \frac{1}{3} 1 - v^{6})]_{0}^{x} : [\frac{1}{3} (v^{3} arcsin (v^{3}) + 1 - v^{6})]_{0}^{x}

= [\frac{1}{3} (v^{3} arcsin (v^{3}) + 1 - v^{6})]_{0}^{x}

Berechne die Grenzen:

\frac{1}{3} (x^{3} arcsin (x^{3}) + 1 - x^{6}) - \frac{1}{3}

= \frac{1}{3} (x^{3} arcsin (x^{3}) + 1 - x^{6}) - \frac{1}{3}

Vereinfache

= \frac{x ^{3} arcsin ( x ^{3} ) + 1 - x ^{6} - 1}{3}

\int_{1}^{6} x^{- 3} d x

\int_{0}^{b} x^{2} d x

\int_{0}^{1} x e^{x^{2}} - x d x

\int_{2}^{6} 15 ln (x + 5) d x

\int_{- 3}^{2} e^{x + 3} d x