integral of ((4x-1))/((2x+1))

Lösung

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

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

Schritte zur Lösung

\int \frac{4 x - 1}{2 x + 1} d x

Wende U-Substitution an

= \int \frac{2 u - 3}{2 u} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int \frac{2 u - 3}{u} d u

Multipliziere aus

\frac{2 u - 3}{u} : 2 - \frac{3}{u}

= \frac{1}{2} \cdot \int 2 - \frac{3}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 2 d u - \int \frac{3}{u} d u)

\int 2 d u = 2 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{2} (2 u - 3 ln ∣ u ∣)

Setze in

u = 2 x + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

\int \frac{y}{y ^{2} - x ^{2}} d y

\int e^{- 0.002 x} d x

\int 4 sin (6 t) d t

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