integral of ((4x-3))/(x^2+7x+12)

Lösung

\int \frac{( 4 x - 3 )}{x ^{2} + 7 x + 12} d x

2 ln 4 x^{2} + 28 x + 48 + 17 ln ∣ 2 x + 8 ∣ - 17 ln ∣ 2 x + 6 ∣ + C

Schritte zur Lösung

\int \frac{4 x - 3}{x ^{2} + 7 x + 12} d x

Vervollständige das Quadrat

x^{2} + 7 x + 12 : (x + \frac{7}{2})^{2} - \frac{1}{4}

= \int \frac{4 x - 3}{( x + \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

Wende U-Substitution an

= \int \frac{4 ( 4 u - 17 )}{4 u ^{2} - 1} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{4 u - 17}{4 u ^{2} - 1} d u

Multipliziere aus

\frac{4 u - 17}{4 u ^{2} - 1} : \frac{4 u}{4 u ^{2} - 1} - \frac{17}{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

= 4 (\int \frac{4 u}{4 u ^{2} - 1} d u - \int \frac{17}{4 u ^{2} - 1} d u)

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

\int \frac{17}{4 u ^{2} - 1} d u = - \frac{17}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 4 (\frac{1}{2} ln 4 u^{2} - 1 - (- \frac{17}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)))

Setze in

u = x + \frac{7}{2}

ein

= 4 (\frac{1}{2} ln 4 (x + \frac{7}{2})^{2} - 1 - (- \frac{17}{4} (ln 2 (x + \frac{7}{2}) + 1 - ln 2 (x + \frac{7}{2}) - 1)))

Vereinfache

4 (\frac{1}{2} ln 4 (x + \frac{7}{2})^{2} - 1 - (- \frac{17}{4} (ln 2 (x + \frac{7}{2}) + 1 - ln 2 (x + \frac{7}{2}) - 1))) : 2 ln 4 x^{2} + 28 x + 48 + 17 ln ∣ 2 x + 8 ∣ - 17 ln ∣ 2 x + 6 ∣

= 2 ln 4 x^{2} + 28 x + 48 + 17 ln ∣ 2 x + 8 ∣ - 17 ln ∣ 2 x + 6 ∣

Füge eine Konstante zur Lösung hinzu

= 2 ln 4 x^{2} + 28 x + 48 + 17 ln ∣ 2 x + 8 ∣ - 17 ln ∣ 2 x + 6 ∣ + C

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