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

Lösung

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

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

Schritte zur Lösung

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

Wende integrale Substitution an

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

Multipliziere aus

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

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

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

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

Setze in

u = \frac{x}{2}

ein

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

(\frac{x}{2})^{2} = \frac{x ^{2}}{4}

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial x} (7 x^{3 y})

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

d er i v a t i v e (\frac{( e ^{6} )}{( e ^{4} )})^{- 1}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 4})

x \frac{d v}{d x} = \frac{1 - 4 v ^{2}}{3 v}