integral of (x^3-1)/(x^3+x)

Lösung

\int \frac{x ^{3} - 1}{x ^{3} + x} d x

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

Schritte zur Lösung

\int \frac{x ^{3} - 1}{x ^{3} + x} d x

\frac{x ^{3} - 1}{x ^{3} + x} = \frac{- x - 1}{x ^{3} + x} + 1

= \int \frac{- x - 1}{x ^{3} + x} + 1 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{- x - 1}{x ^{3} + x} d x + \int 1 d x

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{x - 1 \cdot ( x ^{2} )} d x

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

\int (4 + x) d x

t a y l or e^{\frac{1}{θ}}

n = 0 \sum \infty \frac{3}{n n}