integral of (x+9)/(x^2+9)

Lösung

\int \frac{x + 9}{x ^{2} + 9} d x

\frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3}) + C

Schritte zur Lösung

\int \frac{x + 9}{x ^{2} + 9} d x

Wende integrale Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = \frac{x}{3}

ein

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

(\frac{x}{3})^{2} = \frac{x ^{2}}{9}

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3}) + C

\frac{\partial}{\partial x} ((x^{2} - 3 y) (x - 2))

\frac{\partial}{\partial y} (\frac{e ^{x}}{e ^{x} + e ^{y}})

\frac{\partial}{\partial x} (5 x^{3} y + 3 sin (x y))

\int (2^{x} - 2)^{2} d x

\int x \cdot cos (x^{2} + 1) d x