integral of (x+3)/(x^2+2x+5)

Lösung

\int \frac{x + 3}{x ^{2} + 2 x + 5} d x

\frac{1}{2} ln \frac{1}{4} (x + 1)^{2} + 1 + arctan (\frac{1}{2} (x + 1)) + C

Schritte zur Lösung

\int \frac{x + 3}{x ^{2} + 2 x + 5} d x

Vervollständige das Quadrat

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Multipliziere aus

\frac{v + 1}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{1}{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

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

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

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

= \frac{1}{2} ln v^{2} + 1 + arctan (v)

Ersetze zurück

= \frac{1}{2} ln (\frac{x + 1}{2})^{2} + 1 + arctan (\frac{x + 1}{2})

Vereinfache

\frac{1}{2} ln (\frac{x + 1}{2})^{2} + 1 + arctan (\frac{x + 1}{2}) : \frac{1}{2} ln \frac{1}{4} (x + 1)^{2} + 1 + arctan (\frac{1}{2} (x + 1))

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

Füge eine Konstante zur Lösung hinzu

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

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