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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \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

= \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)

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

Setze in

u = x - 2

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d y}{d x} + y = - 1, y (0) = 1

d er i v a t i v e 2 x - 3

t an g e n t f (x) = 3 x + 4, at x = 4

t an g e n t f (x) = x^{4} - 32 x^{2} + 256, at (1,)

\int cos (θ) d θ