integral of (6x+4)/(3x^2+4x+2)

Lösung

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 2} d x

ln 9 x^{2} + 12 x + 6 + C

Schritte zur Lösung

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 2} d x

Multipliziere aus

\frac{6 x + 4}{3 x ^{2} + 4 x + 2} : \frac{6 x}{3 x ^{2} + 4 x + 2} + \frac{4}{3 x ^{2} + 4 x + 2}

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}{3 x ^{2} + 4 x + 2} d x + \int \frac{4}{3 x ^{2} + 4 x + 2} d x

\int \frac{6 x}{3 x ^{2} + 4 x + 2} d x = ln 9 x^{2} + 12 x + 6 - 22 arctan (\frac{3}{2} x + 2)

\int \frac{4}{3 x ^{2} + 4 x + 2} d x = 22 arctan (\frac{3}{2} x + 2)

= ln 9 x^{2} + 12 x + 6 - 22 arctan (\frac{3}{2} x + 2) + 22 arctan (\frac{3}{2} x + 2)

Vereinfache

= ln 9 x^{2} + 12 x + 6

Füge eine Konstante zur Lösung hinzu

= ln 9 x^{2} + 12 x + 6 + C

\frac{\partial}{\partial x} (x \cdot ln (x + 2 y))

y^{^{'}} = \frac{4 t + y}{4 y - t}

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

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}

\int \frac{1 + cos ( x )}{1 + sin ( x )} d x