integral of 1/(x^2(4+3x))

Lösung

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

3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x}) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

\frac{1}{u ( u - 4 ) ^{2}} : \frac{1}{16 u} - \frac{1}{16 ( u - 4 )} + \frac{1}{4 ( u - 4 ) ^{2}}

= 3 \cdot \int \frac{1}{16 u} - \frac{1}{16 ( u - 4 )} + \frac{1}{4 ( u - 4 ) ^{2}} 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

= 3 (\int \frac{1}{16 u} d u - \int \frac{1}{16 ( u - 4 )} d u + \int \frac{1}{4 ( u - 4 ) ^{2}} d u)

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

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

= 3 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{4 ( u - 4 )})

Setze in

u = 4 + 3 x

ein

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

Vereinfache

3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln ∣ 4 + 3 x - 4 ∣ - \frac{1}{4 ( 4 + 3 x - 4 )}) : 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x})

= 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x})

Füge eine Konstante zur Lösung hinzu

= 3 (\frac{1}{16} ln ∣ 4 + 3 x ∣ - \frac{1}{16} ln (3 ∣ x ∣) - \frac{1}{12 x}) + C

a re a y = 7 x^{2}, y = x^{2} + 3

t an g e n t f (x) = \frac{10 x}{x ^{2} - 6}, at x = 4

\int \frac{6 x ^{4} + 24 x ^{2} + 4}{x ^{3} + 4 x} d x

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

d er i v a t i v e f (x) = 2 x + 3