integral of (x^4)/(x^4-16)

Lösung

\int \frac{x ^{4}}{x ^{4} - 16} d x

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

Schritte zur Lösung

\int \frac{x ^{4}}{x ^{4} - 16} d x

\frac{x ^{4}}{x ^{4} - 16} = \frac{16}{x ^{4} - 16} + 1

= \int \frac{16}{x ^{4} - 16} + 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{16}{x ^{4} - 16} d x + \int 1 d x

\int \frac{16}{x ^{4} - 16} d x = - arctan (\frac{x}{2}) - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 2 ∣

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{''}} + 9 y = 2 cos (3 x) + 3 sin (3 x)

t an g e n t f (x) = 4 sin (x), at x = \frac{π}{2}

y y^{^{'}} = y^{2} + 1

d er i v a t i v e f (x) = csc^{3} (5 x^{2} + 1)

\int cot^{4} (\frac{1}{2}) t d t