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

Lösung

integral

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u + 1}{- 4 u ^{2} + 5} d u

Multipliziere aus

\frac{2 u + 1}{- 4 u ^{2} + 5} : \frac{2 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{2 u}{- 4 u ^{2} + 5} d u = - \frac{1}{4} ln - 4 u^{2} + 5

\int \frac{1}{- 4 u ^{2} + 5} d u = \frac{1}{4 5} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

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

Setze in

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

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

in t e g r a l (1 - x)^{5}

in t e g r a l \frac{1 + 1}{x ^{4}}

in t e g r a l \frac{e ^{x} + 1}{e ^{x}}

in t e g r a l e^{7} y