integral of (x^2)/(x^2-4x+5)

Lösung

\int \frac{x ^{2}}{x ^{2} - 4 x + 5} d x

2 ln x^{2} - 4 x + 5 + 3 arctan (x - 2) + x + C

Schritte zur Lösung

\int \frac{x ^{2}}{x ^{2} - 4 x + 5} d x

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

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

\int \frac{4 x - 5}{x ^{2} - 4 x + 5} d x = 2 ln x^{2} - 4 x + 5 + 3 arctan (x - 2)

\int 1 d x = x

= 2 ln x^{2} - 4 x + 5 + 3 arctan (x - 2) + x

Füge eine Konstante zur Lösung hinzu

= 2 ln x^{2} - 4 x + 5 + 3 arctan (x - 2) + x + C

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

t an g e n t x (49.7)

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

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 7, y^{^{'}} (0) = - 3

x \frac{d y}{d x} + (3 x + 1) y = e^{- 3 x}