integral of (5x^2+20x+6)/(x(x+1)^2)

Lösung

\int \frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} d x

- ln ∣ x + 1 ∣ - \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} d x

Multipliziere aus

\frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} : \frac{5 x ^{2}}{x ( x + 1 ) ^{2}} + \frac{20 x}{x ( x + 1 ) ^{2}} + \frac{6}{x ( x + 1 ) ^{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{5 x ^{2}}{x ( x + 1 ) ^{2}} d x + \int \frac{20 x}{x ( x + 1 ) ^{2}} d x + \int \frac{6}{x ( x + 1 ) ^{2}} d x

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

\int \frac{20 x}{x ( x + 1 ) ^{2}} d x = - \frac{20}{x + 1}

\int \frac{6}{x ( x + 1 ) ^{2}} d x = 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

= 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{20}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Vereinfache

5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{20}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + 6 ln ∣ x ∣

= - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

\int x (2 x - 3) d x

\frac{\partial}{\partial x} (x^{3} - 6 x y + y^{3} - 2)

\frac{d}{d t} (+ 6 t)

\int_{4}^{10} \frac{1}{36 + ( x - 4 ) ^{2}} d x

\frac{d}{d x} (6 x - 6 e^{x})