integral of (2x+1)/(x^2+2x+37)

Lösung

\int \frac{2 x + 1}{x ^{2} + 2 x + 37} d x

ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1)) + C

Schritte zur Lösung

\int \frac{2 x + 1}{x ^{2} + 2 x + 37} d x

Vervollständige das Quadrat

x^{2} + 2 x + 37 : (x + 1)^{2} + 36

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 36} d x

Wende U-Substitution an

= \int \frac{2 u - 1}{u ^{2} + 36} d u

Multipliziere aus

\frac{2 u - 1}{u ^{2} + 36} : \frac{2 u}{u ^{2} + 36} - \frac{1}{u ^{2} + 36}

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{2 u}{u ^{2} + 36} d u - \int \frac{1}{u ^{2} + 36} d u

\int \frac{2 u}{u ^{2} + 36} d u = ln u^{2} + 36

\int \frac{1}{u ^{2} + 36} d u = \frac{1}{6} arctan (\frac{u}{6})

= ln u^{2} + 36 - \frac{1}{6} arctan (\frac{u}{6})

Setze in

u = x + 1

ein

= ln (x + 1)^{2} + 36 - \frac{1}{6} arctan (\frac{x + 1}{6})

Vereinfache

ln (x + 1)^{2} + 36 - \frac{1}{6} arctan (\frac{x + 1}{6}) : ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1))

= ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1))

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1)) + C

\frac{d}{d x} (sin (x) \cdot sin (x))

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