integral of (5x+4)/((x+8)^2(x-1))

Lösung

\int \frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} d x

- \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8} + C

Schritte zur Lösung

\int \frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} d x

Multipliziere aus

\frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} : \frac{5 x}{( x + 8 ) ^{2} ( x - 1 )} + \frac{4}{( x + 8 ) ^{2} ( x - 1 )}

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{5 x}{( x + 8 ) ^{2} ( x - 1 )} d x + \int \frac{4}{( x + 8 ) ^{2} ( x - 1 )} d x

\int \frac{5 x}{( x + 8 ) ^{2} ( x - 1 )} d x = 5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

\int \frac{4}{( x + 8 ) ^{2} ( x - 1 )} d x = 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

= 5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) + 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣)

Vereinfache

5 (- \frac{1}{81} ln ∣ x + 8 ∣ - \frac{8}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) + 4 (- \frac{1}{81} ln ∣ x + 8 ∣ + \frac{1}{9 ( x + 8 )} + \frac{1}{81} ln ∣ x - 1 ∣) : - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8}

= - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{9} ln ∣ x + 8 ∣ + \frac{1}{9} ln ∣ x - 1 ∣ - \frac{4}{x + 8} + C

\frac{d y}{d x} = \frac{x - 1}{x + 3} \frac{y - 2}{y - 3}

in v erse l a pl a ce \frac{2 s + 3}{s ^{2} + 6 s + 13}

\frac{1}{5} \frac{d y}{d x} = (\frac{1}{5})^{2} - \frac{1}{5}

in v erse l a pl a ce \frac{5 x + 5}{x ^{2} - 4}

\frac{\partial}{\partial x} (12 x y^{2})