integral of (3/(x^4)+2-3/(x^2))

Lösung

\int (\frac{3}{x ^{4}} + 2 - \frac{3}{x ^{2}}) d x

- \frac{1}{x ^{3}} + 2 x + \frac{3}{x} + C

Schritte zur Lösung

\int \frac{3}{x ^{4}} + 2 - \frac{3}{x ^{2}} 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{3}{x ^{4}} d x + \int 2 d x - \int \frac{3}{x ^{2}} d x

\int \frac{3}{x ^{4}} d x = - \frac{1}{x ^{3}}

\int 2 d x = 2 x

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

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

Vereinfache

= - \frac{1}{x ^{3}} + 2 x + \frac{3}{x}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{x ^{3}} + 2 x + \frac{3}{x} + C

\int_{1}^{\infty} x e^{- 6 x} d x

\frac{d}{d x} (\frac{4 x - 7}{x - 2})

p a r i t y x^{4} sin (x) tan (x)

t an g e n t f (x) = 5 + 5 x^{2} - 2 x^{3}, at x = 1

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