integral of (4x+5)/(x^2+4x+20)

Lösung

\int \frac{4 x + 5}{x ^{2} + 4 x + 20} d x

\frac{1}{4} (8 ln \frac{1}{16} (x + 2)^{2} + 1 - 3 arctan (\frac{1}{4} (x + 2))) + C

Schritte zur Lösung

\int \frac{4 x + 5}{x ^{2} + 4 x + 20} d x

Vervollständige das Quadrat

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int \frac{4 x + 5}{( x + 2 ) ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{4 u - 3}{u ^{2} + 16} d u

Wende integrale Substitution an

= \int \frac{16 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

= \frac{1}{4} \cdot \int \frac{16 v - 3}{v ^{2} + 1} d v

Multipliziere aus

\frac{16 v - 3}{v ^{2} + 1} : \frac{16 v}{v ^{2} + 1} - \frac{3}{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

= \frac{1}{4} (\int \frac{16 v}{v ^{2} + 1} d v - \int \frac{3}{v ^{2} + 1} d v)

\int \frac{16 v}{v ^{2} + 1} d v = 8 ln v^{2} + 1

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= \frac{1}{4} (8 ln v^{2} + 1 - 3 arctan (v))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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