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

Lösung

integral

\frac{x}{( x ^{2} - 2 x + 2 )}

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{cos ^{23} ( x )}{sin ( x )}

in t e g r a l \frac{3}{4} - \frac{x}{4}

in t e g r a l y = \frac{x}{2 + 1}

in t e g r a l \frac{x}{( 1 + x - x ^{2} )}

in t e g r a l (x + (x^{2} + 1)^{\frac{1}{2}})^{- 1}