integral of 1/(x^2+3x+9)

Lösung

\int \frac{1}{x ^{2} + 3 x + 9} d x

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

Schritte zur Lösung

\int \frac{1}{x ^{2} + 3 x + 9} d x

Vervollständige das Quadrat

x^{2} + 3 x + 9 : (x + \frac{3}{2})^{2} + \frac{27}{4}

= \int \frac{1}{( x + \frac{3}{2} ) ^{2} + \frac{27}{4}} d x

Wende U-Substitution an

= \int \frac{4}{4 u ^{2} + 27} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 27} d u

Wende integrale Substitution an

= 4 \cdot \int \frac{1}{6 3 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 4 \cdot \frac{1}{6 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= 4 \cdot \frac{1}{6 3} arctan (v)

Ersetze zurück

= 4 \cdot \frac{1}{6 3} arctan (\frac{2}{3 3} (x + \frac{3}{2}))

Vereinfache

4 \cdot \frac{1}{6 3} arctan (\frac{2}{3 3} (x + \frac{3}{2})) : \frac{2}{3 3} arctan (\frac{2}{3 3} x + \frac{1}{3})

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

Füge eine Konstante zur Lösung hinzu

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

f (x) = \frac{1}{x + 2}

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

\int \frac{x ^{2}}{x ^{6} + 9} d x

\frac{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

a re a y = x^{2}, y = 2 x