integral of (7x+4)/(x^3+4x^2)

Lösung

\int \frac{7 x + 4}{x ^{3} + 4 x ^{2}} d x

\frac{3}{2} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 4 ∣ - \frac{1}{x} + C

Schritte zur Lösung

\int \frac{7 x + 4}{x ^{3} + 4 x ^{2}} d x

Multipliziere aus

\frac{7 x + 4}{x ^{3} + 4 x ^{2}} : \frac{7 x}{x ^{3} + 4 x ^{2}} + \frac{4}{x ^{3} + 4 x ^{2}}

Wende die Summenregel an:

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

= \int \frac{7 x}{x ^{3} + 4 x ^{2}} d x + \int \frac{4}{x ^{3} + 4 x ^{2}} d x

\int \frac{7 x}{x ^{3} + 4 x ^{2}} d x = 7 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣)

\int \frac{4}{x ^{3} + 4 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

= 7 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

Vereinfache

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

= \frac{3}{2} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 4 ∣ - \frac{1}{x}

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 4 ∣ - \frac{1}{x} + C

y^{^{'}} + 3 (tan (3 x)) y = 7 cos (3 x)

x \to 1 lim ((1 - x)^{l n (x)})

\int_{1}^{5} 4 x^{2} ln (x) d x

f^{^{'}} (x) = 2 sin (6 x), f (\frac{π}{6}) = - \frac{5}{2}

x \to 1 lim (sin (5 p i x))