integral of (7x^2)/((9+x^2)^2)

Lösung

\int \frac{7 x ^{2}}{( 9 + x ^{2} ) ^{2}} d x

\frac{7 ( x ^{2} arctan ( \frac{x}{3} ) - 3 x + 9 arctan ( \frac{x}{3} ) )}{6 ( 9 + x ^{2} )} + C

Schritte zur Lösung

\int \frac{7 x ^{2}}{( 9 + x ^{2} ) ^{2}} d x

Entferne die Konstante:

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

= 7 \cdot \int \frac{x ^{2}}{( 9 + x ^{2} ) ^{2}} d x

Wende integrale Substitution an

= 7 \cdot \int \frac{u ^{2}}{3 ( u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

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

= 7 \cdot \frac{1}{3} \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}}

= 7 \cdot \frac{1}{3} \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

= 7 \cdot \frac{1}{3} (\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}})

= 7 \cdot \frac{1}{3} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Setze in

u = \frac{x}{3}

ein

= 7 \cdot \frac{1}{3} (arctan (\frac{x}{3}) - \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}}))

Vereinfache

7 \cdot \frac{1}{3} (arctan (\frac{x}{3}) - \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}})) : \frac{7 ( x ^{2} arctan ( \frac{x}{3} ) - 3 x + 9 arctan ( \frac{x}{3} ) )}{6 ( 9 + x ^{2} )}

= \frac{7 ( x ^{2} arctan ( \frac{x}{3} ) - 3 x + 9 arctan ( \frac{x}{3} ) )}{6 ( 9 + x ^{2} )}

Füge eine Konstante zur Lösung hinzu

= \frac{7 ( x ^{2} arctan ( \frac{x}{3} ) - 3 x + 9 arctan ( \frac{x}{3} ) )}{6 ( 9 + x ^{2} )} + C

\int_{0}^{1} 16 sin^{4} (x) d x

x \to 3 lim (\frac{x - 3}{( x - 3 )})

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\frac{d}{d x} (e^{\frac{4}{x}})

x \to 2 lim (5 x - 6)