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

Lösung

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

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

(x + 3) (x - 2) : (x + \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 25} 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 - 1}{4 u ^{2} - 25} d u

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} - \frac{1}{4 u ^{2} - 25}

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} - 25} d u - \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

Setze in

u = x + \frac{1}{2}

ein

= 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

Vereinfache

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

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