integral of (11x+20)/(x^2+4x+5)

Lösung

\int \frac{11 x + 20}{x ^{2} + 4 x + 5} d x

\frac{11}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2) + C

Schritte zur Lösung

\int \frac{11 x + 20}{x ^{2} + 4 x + 5} d x

Vervollständige das Quadrat

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

= \int \frac{11 x + 20}{( x + 2 ) ^{2} + 1} d x

Wende U-Substitution an

= \int \frac{11 u - 2}{u ^{2} + 1} d u

Multipliziere aus

\frac{11 u - 2}{u ^{2} + 1} : \frac{11 u}{u ^{2} + 1} - \frac{2}{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

= \int \frac{11 u}{u ^{2} + 1} d u - \int \frac{2}{u ^{2} + 1} d u

\int \frac{11 u}{u ^{2} + 1} d u = \frac{11}{2} ln u^{2} + 1

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

= \frac{11}{2} ln u^{2} + 1 - 2 arctan (u)

Setze in

u = x + 2

ein

= \frac{11}{2} ln (x + 2)^{2} + 1 - 2 arctan (x + 2)

Vereinfache

\frac{11}{2} ln (x + 2)^{2} + 1 - 2 arctan (x + 2) : \frac{11}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)

= \frac{11}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)

Füge eine Konstante zur Lösung hinzu

= \frac{11}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2) + C

f (x) = \frac{x}{4 + x ^{2}}

\frac{\partial}{\partial z} (e^{9 x z})

\int 2 sin (nπ x) d x

\frac{d}{d x} ((x^{2} - x^{\frac{7}{3}})^{4})

10 y^{^{'}} + 5 y = 8 e^{6 x}, y (0) = 9