integral of (x^2)/((64+x^2)^2)

Lösung

\int \frac{x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

\frac{1}{8} (\frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}}) + C

Schritte zur Lösung

\int \frac{x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

Wende integrale Substitution an

= \int \frac{u ^{2}}{8 ( u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Ermittle den Partialbruch von

\frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} : \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}}

= \frac{1}{8} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= \frac{1}{8} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Setze in

u = \frac{x}{8}

ein

= \frac{1}{8} (arctan (\frac{x}{8}) - \frac{1}{2} (arctan (\frac{x}{8}) + \frac{\frac{x}{8}}{1 + ( \frac{x}{8} ) ^{2}}))

Vereinfache

\frac{1}{8} (arctan (\frac{x}{8}) - \frac{1}{2} (arctan (\frac{x}{8}) + \frac{\frac{x}{8}}{1 + ( \frac{x}{8} ) ^{2}})) : \frac{1}{8} (\frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}})

= \frac{1}{8} (\frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (\frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}}) + C

(5 x y^{2} - 3) d x + (5 x^{2} y + 4) d y = 0

(x, y) \to (0, 0) lim (x^{3} y)

s im pl i f y \frac{3}{12} x^{4} + \frac{2}{6} x^{3} - x^{2} + 1

\frac{d}{d x} (\frac{6 x}{( x ^{2} + 3 ) ^{2}})

t an g e n t - \frac{1}{2} x^{2} - 2 x + 3