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

Lösung

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

ln 4 x - 12 x + 8 - 3 ln 2 x - 2 + 3 ln 2 x - 4 + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{2 v + 3}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} + \frac{3}{4 v ^{2} - 1}

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} - 1} d v + \int \frac{3}{4 v ^{2} - 1} d v)

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

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

= 2 \cdot 2 (\frac{1}{4} ln 4 v^{2} - 1 - \frac{3}{4} (ln ∣ 2 v + 1 ∣ - ln ∣ 2 v - 1 ∣))

Ersetze zurück

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

Vereinfache

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

= ln 4 x - 12 x + 8 - 3 ln 2 x - 2 + 3 ln 2 x - 4

Füge eine Konstante zur Lösung hinzu

= ln 4 x - 12 x + 8 - 3 ln 2 x - 2 + 3 ln 2 x - 4 + C

\int e^{- 0.08 t} d t

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