integral of (3+x)/(1-x^2)

Lösung

\int \frac{3 + x}{1 - x ^{2}} d x

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

Schritte zur Lösung

\int \frac{3 + x}{1 - x ^{2}} d x

Multipliziere aus

\frac{3 + x}{1 - x ^{2}} : \frac{3}{1 - x ^{2}} + \frac{x}{1 - x ^{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{3}{1 - x ^{2}} d x + \int \frac{x}{1 - x ^{2}} d x

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

\int \frac{x}{1 - x ^{2}} d x = - \frac{1}{2} ln 1 - x^{2}

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

Füge eine Konstante zur Lösung hinzu

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

y = (5 - x^{2})^{5}

t an g e n t f (x) = \frac{1}{x ^{3}}, at x = 1

\frac{d y}{d x} + y = \frac{e ^{x}}{y}

a re a ln (x), \frac{4}{x}, 1, 3

\frac{d}{d x} (b x e^{x})