integral of (2x+7)/(2x^2+2x+1)

Lösung

\int \frac{2 x + 7}{2 x ^{2} + 2 x + 1} d x

\frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1) + C

Schritte zur Lösung

\int \frac{2 x + 7}{2 x ^{2} + 2 x + 1} d x

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 12 u + 37} d u

Vervollständige das Quadrat

u^{2} - 12 u + 37 : (u - 6)^{2} + 1

= \int \frac{u}{( u - 6 ) ^{2} + 1} d u

Wende U-Substitution an

= \int \frac{v + 6}{v ^{2} + 1} d v

Multipliziere aus

\frac{v + 6}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{6}{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{6}{v ^{2} + 1} d v

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

\int \frac{6}{v ^{2} + 1} d v = 6 arctan (v)

= \frac{1}{2} ln v^{2} + 1 + 6 arctan (v)

Ersetze zurück

= \frac{1}{2} ln (2 x + 7 - 6)^{2} + 1 + 6 arctan (2 x + 7 - 6)

Vereinfache

\frac{1}{2} ln (2 x + 7 - 6)^{2} + 1 + 6 arctan (2 x + 7 - 6) : \frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1)

= \frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1) + C

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