integral of (x^3)/(sqrt((x+1)^2+4))

Lösung

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

\frac{1}{3} (x^{2} + 2 x + 5)^{\frac{3}{2}} - x^{2} + 2 x + 5 + 5 ln (\frac{1}{2} x + x^{2} + 2 x + 5 + 1) - \frac{3}{2} (x + 1) x^{2} + 2 x + 5 + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{( u - 1 ) ^{3}}{u ^{2} + 4} d u

Multipliziere aus

\frac{( u - 1 ) ^{3}}{u ^{2} + 4} : \frac{u ^{3}}{u ^{2} + 4} - \frac{3 u ^{2}}{u ^{2} + 4} + \frac{3 u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4}

= \int \frac{u ^{3}}{u ^{2} + 4} - \frac{3 u ^{2}}{u ^{2} + 4} + \frac{3 u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4} 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 \frac{u ^{3}}{u ^{2} + 4} d u - \int \frac{3 u ^{2}}{u ^{2} + 4} d u + \int \frac{3 u}{u ^{2} + 4} d u - \int \frac{1}{u ^{2} + 4} d u

\int \frac{u ^{3}}{u ^{2} + 4} d u = \frac{1}{3} (u^{2} + 4)^{\frac{3}{2}} - 4 u^{2} + 4

\int \frac{3 u ^{2}}{u ^{2} + 4} d u = 12 (- \frac{1}{2} ln (\frac{u + 4 + u ^{2}}{2}) + \frac{1}{8} u 4 + u^{2})

\int \frac{3 u}{u ^{2} + 4} d u = 3 u^{2} + 4

\int \frac{1}{u ^{2} + 4} d u = ln (\frac{1}{2} u + 4 + u^{2})

= \frac{1}{3} (u^{2} + 4)^{\frac{3}{2}} - 4 u^{2} + 4 - 12 (- \frac{1}{2} ln (\frac{u + 4 + u ^{2}}{2}) + \frac{1}{8} u 4 + u^{2}) + 3 u^{2} + 4 - ln (\frac{1}{2} u + 4 + u^{2})

Setze in

u = x + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (x^{2} + 2 x + 5)^{\frac{3}{2}} - x^{2} + 2 x + 5 + 5 ln (\frac{1}{2} x + x^{2} + 2 x + 5 + 1) - \frac{3}{2} (x + 1) x^{2} + 2 x + 5 + C

d er i v a t i v e 1 6^{4} 4 x^{4} + 4

6 y^{^{'}} = x^{2}

\frac{d}{d x} (\frac{1}{3} x^{3} - x^{2} - 8 x + 1)

y^{^{'}} = \frac{2 x ^{4} + y ^{4}}{x y ^{3}}

2 x \frac{d y}{d x} = y^{2}