integral of ((x^3+10))/(x^3-2x^2+5x)

Lösung

\int \frac{( x ^{3} + 10 )}{x ^{3} - 2 x ^{2} + 5 x} d x

x - \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + ln \frac{1}{4} (x - 1)^{2} + 1 + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5 + C

Schritte zur Lösung

\int \frac{x ^{3} + 10}{x ^{3} - 2 x ^{2} + 5 x} d x

Multipliziere aus

\frac{x ^{3} + 10}{x ^{3} - 2 x ^{2} + 5 x} : \frac{x ^{3}}{x ^{3} - 2 x ^{2} + 5 x} + \frac{10}{x ^{3} - 2 x ^{2} + 5 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{x ^{3}}{x ^{3} - 2 x ^{2} + 5 x} d x + \int \frac{10}{x ^{3} - 2 x ^{2} + 5 x} d x

\int \frac{x ^{3}}{x ^{3} - 2 x ^{2} + 5 x} d x = ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x

\int \frac{10}{x ^{3} - 2 x ^{2} + 5 x} d x = 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1))

= ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1))

Vereinfache

ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) : x - \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + ln \frac{1}{4} (x - 1)^{2} + 1 + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5

= x - \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + ln \frac{1}{4} (x - 1)^{2} + 1 + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5

Füge eine Konstante zur Lösung hinzu

= x - \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + ln \frac{1}{4} (x - 1)^{2} + 1 + 2 ln ∣ x ∣ - ln x^{2} - 2 x + 5 + C

\int - cos (y) d y

n = 0 \sum \infty 4 \cdot (- \frac{1}{8})^{n}

y \to - 2 lim (\frac{y ^{3} + 8}{y + 2})

\int cos (x) sec (x) tan (x) d x

\int_{0}^{2} 2 x - 1 d x