integral of (4x^2)/((16+x^2)^2)

Lösung

\int \frac{4 x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

\frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}} + C

Schritte zur Lösung

\int \frac{4 x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

Entferne die Konstante:

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

= 4 \cdot \int \frac{x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

Wende integrale Substitution an

= 4 \cdot \int \frac{u ^{2}}{4 ( u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

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

= 4 \cdot \frac{1}{4} \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}}

= 4 \cdot \frac{1}{4} \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

= 4 \cdot \frac{1}{4} (\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}})

= 4 \cdot \frac{1}{4} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Setze in

u = \frac{x}{4}

ein

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

Vereinfache

4 \cdot \frac{1}{4} (arctan (\frac{x}{4}) - \frac{1}{2} (arctan (\frac{x}{4}) + \frac{\frac{x}{4}}{1 + ( \frac{x}{4} ) ^{2}})) : \frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}}

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}} + C

y^{^{''}} - 6 y^{^{'}} - 7 y = 15 e^{2 t}

\int - \frac{5}{x ^{2}} d x

d er i v a t i v e \frac{4 4 q}{1 - q}

\frac{d y}{d x} + y = 2 x

\frac{\partial}{\partial y} (\frac{4 y}{x ^{5}})