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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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{1}{16 x} d x - \int \frac{1}{16 ( x + 4 )} d x - \int \frac{1}{4 ( x + 4 ) ^{2}} d x

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

\int \frac{1}{16 ( x + 4 )} d x = \frac{1}{16} ln ∣ x + 4 ∣

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int_{1}^{3} 3 x (x - 4) d x

im pl i c i t \frac{d y}{d x}, e^{4 x} = sin (x + 5 y)

d er i v a t i v e \frac{e ^{x}}{1 + x}

\int \frac{sin ( x )}{1 + cos ^{2} ( x )} d x

\frac{d}{d x} (x^{3} - 7 x^{2})