integral of xsqrt(x^2+8x+21)

Lösung

\int x x^{2} + 8 x + 21 d x

\frac{1}{3} (x^{2} + 8 x + 21)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) x^{2} + 8 x + 21 + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{x ^{2} + 8 x + 21}{5}) + C

Schritte zur Lösung

\int x x^{2} + 8 x + 21 d x

Vervollständige das Quadrat

x^{2} + 8 x + 21 : (x + 4)^{2} + 5

= \int x (x + 4)^{2} + 5 d x

Wende U-Substitution an

= \int (u - 4) u^{2} + 5 d u

Multipliziere aus

(u - 4) u^{2} + 5 : u u^{2} + 5 - 4 u^{2} + 5

= \int u u^{2} + 5 - 4 u^{2} + 5 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 u^{2} + 5 d u - \int 4 u^{2} + 5 d u

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

\int 4 u^{2} + 5 d u = 20 (\frac{1}{10} u 5 + u^{2} + \frac{1}{2} ln \frac{1}{5} u + \frac{1}{5} (5 + u^{2}))

= \frac{1}{3} (u^{2} + 5)^{\frac{3}{2}} - 20 (\frac{1}{10} u 5 + u^{2} + \frac{1}{2} ln \frac{1}{5} u + \frac{1}{5} (5 + u^{2}))

Setze in

u = x + 4

ein

= \frac{1}{3} ((x + 4)^{2} + 5)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) 5 + (x + 4)^{2} + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{1}{5} (5 + (x + 4)^{2}))

Vereinfache

\frac{1}{3} ((x + 4)^{2} + 5)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) 5 + (x + 4)^{2} + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{1}{5} (5 + (x + 4)^{2})) : \frac{1}{3} (x^{2} + 8 x + 21)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) x^{2} + 8 x + 21 + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{x ^{2} + 8 x + 21}{5})

= \frac{1}{3} (x^{2} + 8 x + 21)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) x^{2} + 8 x + 21 + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{x ^{2} + 8 x + 21}{5})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (x^{2} + 8 x + 21)^{\frac{3}{2}} - 20 (\frac{1}{10} (x + 4) x^{2} + 8 x + 21 + \frac{1}{2} ln \frac{1}{5} (x + 4) + \frac{x ^{2} + 8 x + 21}{5}) + C

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