integral of x/(x^2-4x+7)

Lösung

\int \frac{x}{x ^{2} - 4 x + 7} d x

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

Schritte zur Lösung

\int \frac{x}{x ^{2} - 4 x + 7} d x

Vervollständige das Quadrat

x^{2} - 4 x + 7 : (x - 2)^{2} + 3

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

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

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

n = 0 \sum \infty 3^{n} x^{n}

- 2 x^{\frac{3}{2}}

d er i v a t i v e y = 9 x^{3}

y^{^{'}} = \frac{x - 4}{y - 6}

s l o p e in t erce pt (4, 5), (1, 2)