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

Lösung

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

\frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \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

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

Multipliziere aus

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

= \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

= \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 ∣

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

Setze in

u = 2 x - 1

ein

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

Vereinfache

\frac{1}{8} (\frac{( 2 x - 1 ) ^{2}}{2} + 2 (2 x - 1) + ln ∣ 2 x - 1 ∣) : \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3)

= \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + C

\int x (4 - x) d x

t an g e n t f (x) = (\frac{x}{1 + x})^{5}, at x = 1

l a pl a ce t r an s f or m 8^{t} \cdot t + 10

\frac{d}{d y} (tan (x^{2} + y))

d er i v a t i v e x^{3} - 4