integral of (x^2)/(x^2-6x+25)

Lösung

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C

Schritte zur Lösung

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{x ^{2}}{x ^{2} - 6 x + 25} = \frac{6 x - 25}{x ^{2} - 6 x + 25} + 1

= \int \frac{6 x - 25}{x ^{2} - 6 x + 25} + 1 d x

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{6 x - 25}{x ^{2} - 6 x + 25} d x + \int 1 d x

\int \frac{6 x - 25}{x ^{2} - 6 x + 25} d x = \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3)))

\int 1 d x = x

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C

\frac{\partial}{\partial x} (x^{x})

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

t a y l or \frac{1}{x} lo g_{10} (1 + x)

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