integral of (20x^3-40x^2+7)/(x^2-2x)

Lösung

\int \frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} d x

10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

Schritte zur Lösung

\int \frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} d x

Lange Division

\frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} : 20 x + \frac{7}{x ^{2} - 2 x}

= \int 20 x + \frac{7}{x ^{2} - 2 x} 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 20 x d x + \int \frac{7}{x ^{2} - 2 x} d x

\int 20 x d x = 10 x^{2}

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

= 10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Füge eine Konstante zur Lösung hinzu

= 10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

\int cos (ln (5 x)) d x

\int \frac{4}{15} x^{\frac{5}{2}} + 9 cos (x) + C x + C d x

\int \frac{1}{x ^{2}} + \frac{1}{x} d x

\frac{d}{d x} (ln (ln (6 x)))

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