integral of xsqrt(x^2+6x+13)

Lösung

\int x x^{2} + 6 x + 13 d x

\frac{1}{3} (x^{2} + 6 x + 13)^{\frac{3}{2}} - 12 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2})) + C

Schritte zur Lösung

\int x x^{2} + 6 x + 13 d x

Vervollständige das Quadrat

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = x + 3

ein

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

Vereinfache

\frac{1}{3} ((x + 3)^{2} + 4)^{\frac{3}{2}} - 12 \frac{1}{8} (x + 3) 4 + (x + 3)^{2} + \frac{1}{2} ln \frac{x + 3 + 4 + ( x + 3 ) ^{2}}{2} : \frac{1}{3} (x^{2} + 6 x + 13)^{\frac{3}{2}} - 12 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2}))

= \frac{1}{3} (x^{2} + 6 x + 13)^{\frac{3}{2}} - 12 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (x^{2} + 6 x + 13)^{\frac{3}{2}} - 12 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2})) + C

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