integral of (x^2)/(xsqrt(x^2-x+1))

Lösung

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

x^{2} - x + 1 + \frac{1}{2} ln \frac{2}{3} x - \frac{1}{3} + \frac{4 x ^{2} - 4 x + 4}{3} + C

Schritte zur Lösung

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

Streiche die gemeinsamen Faktoren:

x

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

Vervollständige das Quadrat

x^{2} - x + 1 : (x - \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{x}{( x - \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Wende U-Substitution an

= \int \frac{2 u + 1}{4 u ^{2} + 3} d u

Multipliziere aus

\frac{2 u + 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{2} + 3}

= \int \frac{2 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{2} + 3} 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{2 u}{4 u ^{2} + 3} d u + \int \frac{1}{4 u ^{2} + 3} d u

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{2} 4 u^{2} + 3

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2} ln \frac{2}{3} u + \frac{1}{3} (3 + 4 u^{2})

= \frac{1}{2} 4 u^{2} + 3 + \frac{1}{2} ln \frac{2}{3} u + \frac{1}{3} (3 + 4 u^{2})

Setze in

u = x - \frac{1}{2}

ein

= \frac{1}{2} 4 (x - \frac{1}{2})^{2} + 3 + \frac{1}{2} ln \frac{2}{3} (x - \frac{1}{2}) + \frac{1}{3} (3 + 4 (x - \frac{1}{2})^{2})

Vereinfache

\frac{1}{2} 4 (x - \frac{1}{2})^{2} + 3 + \frac{1}{2} ln \frac{2}{3} (x - \frac{1}{2}) + \frac{1}{3} (3 + 4 (x - \frac{1}{2})^{2}) : x^{2} - x + 1 + \frac{1}{2} ln \frac{2}{3} x - \frac{1}{3} + \frac{4 x ^{2} - 4 x + 4}{3}

= x^{2} - x + 1 + \frac{1}{2} ln \frac{2}{3} x - \frac{1}{3} + \frac{4 x ^{2} - 4 x + 4}{3}

Füge eine Konstante zur Lösung hinzu

= x^{2} - x + 1 + \frac{1}{2} ln \frac{2}{3} x - \frac{1}{3} + \frac{4 x ^{2} - 4 x + 4}{3} + C

\int x cos (\frac{1}{2} x) d x

t an g e n t y = sin (x y) + π, (1, π)

7 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 9 y = 0

t an g e n t f (x) = x^{4} + 3 x^{2} - x, (1, 3)

\int_{0}^{a} x d x