integral of ((6x-1))/(x^3(2x-1))

Lösung

\int \frac{( 6 x - 1 )}{x ^{3} ( 2 x - 1 )} d x

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

Schritte zur Lösung

\int \frac{6 x - 1}{x ^{3} ( 2 x - 1 )} d x

Wende U-Substitution an

= \int \frac{4 ( 3 u + 2 )}{u ( u + 1 ) ^{3}} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{3 u + 2}{u ( u + 1 ) ^{3}} d u

Ermittle den Partialbruch von

\frac{3 u + 2}{u ( u + 1 ) ^{3}} : \frac{2}{u} - \frac{2}{u + 1} - \frac{2}{( u + 1 ) ^{2}} + \frac{1}{( u + 1 ) ^{3}}

= 4 \cdot \int \frac{2}{u} - \frac{2}{u + 1} - \frac{2}{( u + 1 ) ^{2}} + \frac{1}{( u + 1 ) ^{3}} 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

= 4 (\int \frac{2}{u} d u - \int \frac{2}{u + 1} d u - \int \frac{2}{( u + 1 ) ^{2}} d u + \int \frac{1}{( u + 1 ) ^{3}} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

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

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

= 4 (2 ln ∣ u ∣ - 2 ln ∣ u + 1 ∣ - (- \frac{2}{u + 1}) - \frac{1}{2 ( u + 1 ) ^{2}})

Setze in

u = 2 x - 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e 4 a x - x

f^{^{'}} (x) = 0

x \to 2 + lim (2 x + 2 + x)

\int e^{- 0.6 x} d x

t an g e n t y = 9 x^{2} + 3 x - 9