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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{- v + 3}{v ^{2} + 1} d v

Multipliziere aus

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{v ^{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

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= 2 \cdot \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

5 y^{^{''}} + 5 y^{^{'}} - 10 y = 0

\int_{- \infty}^{0} (4 - 3 x) e^{x} d x

\int_{0}^{4} \frac{3}{x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (5 e^{x})

t an g e n t f (x) = 5 x^{2} - 4 x + 3