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

Lösung

\int \frac{1}{x + 2 x + 3} d x

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

Schritte zur Lösung

\int \frac{1}{x + 2 x + 3} d x

Wende U-Substitution an

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= 2 \cdot \int \frac{v - 1}{v ^{2} - 4} d v

Multipliziere aus

\frac{v - 1}{v ^{2} - 4} : \frac{v}{v ^{2} - 4} - \frac{1}{v ^{2} - 4}

Wende die Summenregel an:

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

= 2 (\int \frac{v}{v ^{2} - 4} d v - \int \frac{1}{v ^{2} - 4} d v)

\int \frac{v}{v ^{2} - 4} d v = \frac{1}{2} ln v^{2} - 4

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

= 2 (\frac{1}{2} ln v^{2} - 4 - (- \frac{1}{4} (ln \frac{v}{2} + 1 - ln \frac{v}{2} - 1)))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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