de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

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

Lösung

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

Lösung

\frac{1}{8} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int \frac{x ^{2}}{2 x + 1} d x

Wende U-Substitution an

= 2 \cdot \int \frac{( u - 1 ) ^{2}}{8 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

Multipliziere aus

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= 2 \cdot \frac{1}{8} \cdot \int u - 2 + \frac{1}{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

= 2 \cdot \frac{1}{8} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{8} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Setze in

u = 2 x + 1

ein

= 2 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣)

Vereinfache

2 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣) : \frac{1}{8} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

= \frac{1}{8} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) + C

Graph

Beliebte Beispiele

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

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

derivative of (x+1/(x-2))

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

(dy)/(dx)=sqrt(7y)e^{x+2}

\frac{d y}{d x} = 7 y e^{x + 2}

(\partial)/(\partial x)((x^2+2)(x^3+1))

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

limit as x approaches-5+of (-1)/(x^2-25)

x \to - 5 + lim (\frac{- 1}{x ^{2} - 25})