integral ((2x+1))/((3x+2))

Lösung

integral

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

\frac{2}{3} x + \frac{4}{9} - \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x + 1}{3 x + 2} : \frac{2 x}{3 x + 2} + \frac{1}{3 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{2 x}{3 x + 2} d x + \int \frac{1}{3 x + 2} d x

\int \frac{2 x}{3 x + 2} d x = \frac{2}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣)

\int \frac{1}{3 x + 2} d x = \frac{1}{3} ln ∣ 3 x + 2 ∣

= \frac{2}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣

Vereinfache

\frac{2}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣ : \frac{2}{3} x + \frac{4}{9} - \frac{1}{9} ln ∣ 3 x + 2 ∣

= \frac{2}{3} x + \frac{4}{9} - \frac{1}{9} ln ∣ 3 x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{2}{3} x + \frac{4}{9} - \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

in t e g r a l \frac{1}{x ^{y}}

in t e g r a l 3 e^{2} x

in t e g r a l \frac{2 x}{e ^{x}}

in t e g r a l \frac{x ^{2}}{( x ^{3} + 1 ) ^{\frac{1}{2}}}

in t e g r a l \frac{- 2}{e ^{x}}