integral of (x-1)/((x^2+4x+5)(x+1))

Lösung

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ( x + 1 )} d x

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

Schritte zur Lösung

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ( x + 1 )} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int \frac{u - 3}{( u ^{2} + 1 ) ( u - 1 )} d u

Ermittle den Partialbruch von

\frac{u - 3}{( u ^{2} + 1 ) ( u - 1 )} : \frac{u + 2}{u ^{2} + 1} - \frac{1}{u - 1}

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

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

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

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

Setze in

u = x + 2

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d y}{d x} + x^{2} y = 10 x^{2}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 cos (4 t)

\int_{0}^{0.6} 250 x d x

\int - 2 sin (2 t) d t

d er i v a t i v e ln (x^{61})