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Beliebt Rechnen >

integral of (12x^2+16x+3)/(4x^3+8x^2+3x)

Lösung

\int \frac{12 x ^{2} + 16 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

Lösung

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

Schritte zur Lösung

\int \frac{12 x ^{2} + 16 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

Multipliziere aus

\frac{12 x ^{2} + 16 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} : \frac{12 x ^{2}}{4 x ^{3} + 8 x ^{2} + 3 x} + \frac{16 x}{4 x ^{3} + 8 x ^{2} + 3 x} + \frac{3}{4 x ^{3} + 8 x ^{2} + 3 x}

Wende die Summenregel an:

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

= \int \frac{12 x ^{2}}{4 x ^{3} + 8 x ^{2} + 3 x} d x + \int \frac{16 x}{4 x ^{3} + 8 x ^{2} + 3 x} d x + \int \frac{3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

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

\int \frac{16 x}{4 x ^{3} + 8 x ^{2} + 3 x} d x = 16 (\frac{1}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x + 3 ∣)

\int \frac{3}{4 x ^{3} + 8 x ^{2} + 3 x} d x = ln ∣ x ∣ - \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x + 3 ∣

= 12 (\frac{1}{8} ln 4 x^{2} + 8 x + 3 + \frac{1}{4} (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 1 ∣)) + 16 (\frac{1}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x + 3 ∣) + ln ∣ x ∣ - \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x + 3 ∣

Vereinfache

12 (\frac{1}{8} ln 4 x^{2} + 8 x + 3 + \frac{1}{4} (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 1 ∣)) + 16 (\frac{1}{4} ln ∣ 2 x + 1 ∣ - \frac{1}{4} ln ∣ 2 x + 3 ∣) + ln ∣ x ∣ - \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x + 3 ∣ : \frac{3}{2} ln 4 x^{2} + 8 x + 3 - \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x + 3 ∣ + ln ∣ x ∣

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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