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

Lösung

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

ln 4 x + 4 x + 1 + \frac{1}{5} (ln \frac{2 x + 1 + 1}{5} + 1 - ln \frac{2 x + 1 + 1}{5} - 1) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Vervollständige das Quadrat

u^{2} - 1 + u : (u + \frac{1}{2})^{2} - \frac{5}{4}

= 2 \cdot \int \frac{u}{( u + \frac{1}{2} ) ^{2} - \frac{5}{4}} d u

Wende U-Substitution an

= 2 \cdot \int \frac{2 ( 2 v - 1 )}{4 v ^{2} - 5} d v

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \int \frac{2 v - 1}{4 v ^{2} - 5} d v

Multipliziere aus

\frac{2 v - 1}{4 v ^{2} - 5} : \frac{2 v}{4 v ^{2} - 5} - \frac{1}{4 v ^{2} - 5}

Wende die Summenregel an:

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

= 2 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 5} d v - \int \frac{1}{4 v ^{2} - 5} d v)

\int \frac{2 v}{4 v ^{2} - 5} d v = \frac{1}{4} ln 4 v^{2} - 5

\int \frac{1}{4 v ^{2} - 5} d v = - \frac{1}{4 5} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)

= 2 \cdot 2 (\frac{1}{4} ln 4 v^{2} - 5 - (- \frac{1}{4 5} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)))

Ersetze zurück

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

Vereinfache

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

= ln 4 x + 4 x + 1 + \frac{1}{5} (ln \frac{2 x + 1 + 1}{5} + 1 - ln \frac{2 x + 1 + 1}{5} - 1)

Füge eine Konstante zur Lösung hinzu

= ln 4 x + 4 x + 1 + \frac{1}{5} (ln \frac{2 x + 1 + 1}{5} + 1 - ln \frac{2 x + 1 + 1}{5} - 1) + C

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