integral of (x+3)/(x^2+4x+5)

Lösung

\int \frac{x + 3}{x ^{2} + 4 x + 5} d x

\frac{1}{2} ln x^{2} + 4 x + 5 + arctan (x + 2) + C

Schritte zur Lösung

\int \frac{x + 3}{x ^{2} + 4 x + 5} d x

Wende U-Substitution an

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

Vervollständige das Quadrat

u^{2} - 2 u + 2 : (u - 1)^{2} + 1

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

Wende U-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 (x + 3 - 1)^{2} + 1 + arctan (x + 3 - 1)

Vereinfache

\frac{1}{2} ln (x + 3 - 1)^{2} + 1 + arctan (x + 3 - 1) : \frac{1}{2} ln x^{2} + 4 x + 5 + arctan (x + 2)

= \frac{1}{2} ln x^{2} + 4 x + 5 + arctan (x + 2)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 4 x + 5 + arctan (x + 2) + C

in t e g r a l \frac{1}{x 1 - x} d x

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