integral of x/((x^2-3x+2))

Lösung

\int \frac{x}{( x ^{2} - 3 x + 2 )} d x

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

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

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

Setze in

u = x - \frac{3}{2}

ein

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x ^{5} e ^{x} - 6 x ^{4}}{x ^{5}} d x

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

\int sin (a x) \cdot e^{b x} d x

\int (cos (a x) + sin (a x))^{2} d x

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