integral of (x^3)/(sqrt(x^2+36))

Lösung

\int \frac{x ^{3}}{x ^{2} + 36} d x

\frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36 + C

Schritte zur Lösung

\int \frac{x ^{3}}{x ^{2} + 36} d x

Wende U-Substitution an

= \int u^{2} - 36 d u

Wende die Summenregel an:

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

= \int u^{2} d u - \int 36 d u

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

\int 36 d u = 36 u

= \frac{u ^{3}}{3} - 36 u

Setze in

u = x^{2} + 36

ein

= \frac{( x ^{2} + 36 ) ^{3}}{3} - 36 x^{2} + 36

Vereinfache

\frac{( x ^{2} + 36 ) ^{3}}{3} - 36 x^{2} + 36 : \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36

= \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (x^{2} + 36)^{\frac{3}{2}} - 36 x^{2} + 36 + C

\frac{d}{d x} (- \frac{1}{2 ( x + 2 ) ^{\frac{3}{2}}})

\frac{\partial}{\partial x} (x^{2} + y^{2} - z)

\int 3 - \frac{1}{3} \cdot (x - 4)^{2} d x

\int_{0}^{2} \frac{x ^{2} - 2}{x + 1} d x

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