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

Lösung

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

ln x^{2} + 2 x + 2 - 2 arctan (x + 1) + C

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{x ^{2} + 2 x + 2} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = x + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 2 x + 2 - 2 arctan (x + 1) + C

\int_{0}^{- 8} (- 4 3 x) d x

\int \frac{5 x}{( x + 3 ) ( x - 2 )} d x

\int u^{- 5} (5 - u^{5} + u^{10}) d u

\int x^{18} d x

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 5