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integral of 3/((2x^2+10x+4)^{7/2)}

Lösung

\int \frac{3}{( 2 x ^{2} + 10 x + 4 ) ^{\frac{7}{2}}} d x

\frac{( 2 x + 5 ) ( 40 ( 2 x + 5 ) ^{2} ( x ^{2} + 5 x + 2 ) - 3 ( 2 x + 5 ) ^{4} - 240 ( x ^{2} + 5 x + 2 ) ^{2} )}{98260 \cdot 2 ^{\frac{1}{2}} ( x ^{2} + 5 x + 2 ) ^{\frac{5}{2}}} + C

Schritte zur Lösung

\int \frac{3}{( 2 x ^{2} + 10 x + 4 ) ^{\frac{7}{2}}} d x

Entferne die Konstante:

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

= 3 \cdot \int \frac{1}{( 2 x ^{2} + 10 x + 4 ) ^{\frac{7}{2}}} d x

Vervollständige das Quadrat

2 x^{2} + 10 x + 4 : 2 (x + \frac{5}{2})^{2} - \frac{17}{2}

= 3 \cdot \int \frac{1}{( 2 ( x + \frac{5}{2} ) ^{2} - \frac{17}{2} ) ^{\frac{7}{2}}} d x

Wende U-Substitution an

= 3 \cdot \int \frac{8 \cdot 2 ^{\frac{1}{2}}}{( 4 u ^{2} - 17 ) ^{\frac{7}{2}}} d u

Entferne die Konstante:

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

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \int \frac{1}{( 4 u ^{2} - 17 ) ^{\frac{7}{2}}} d u

Trigonometrische Substitution anwenden

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \int \frac{sec ( v )}{9826 tan ^{6} ( v )} d v

Entferne die Konstante:

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

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} \cdot \int \frac{sec ( v )}{tan ^{6} ( v )} d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} \cdot \int \frac{( 1 - sin ^{2} ( v ) ) ^{2} cos ( v )}{sin ^{6} ( v )} d v

Wende U-Substitution an

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} \cdot \int \frac{( 1 - w ^{2} ) ^{2}}{w ^{6}} d w

Multipliziere aus

\frac{( 1 - w ^{2} ) ^{2}}{w ^{6}} : \frac{1}{w ^{6}} - \frac{2}{w ^{4}} + \frac{1}{w ^{2}}

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} \cdot \int \frac{1}{w ^{6}} - \frac{2}{w ^{4}} + \frac{1}{w ^{2}} d w

Wende die Summenregel an:

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

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} (\int \frac{1}{w ^{6}} d w - \int \frac{2}{w ^{4}} d w + \int \frac{1}{w ^{2}} d w)

\int \frac{1}{w ^{6}} d w = - \frac{1}{5 w ^{5}}

\int \frac{2}{w ^{4}} d w = - \frac{2}{3 w ^{3}}

\int \frac{1}{w ^{2}} d w = - \frac{1}{w}

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} (- \frac{1}{5 w ^{5}} - (- \frac{2}{3 w ^{3}}) - \frac{1}{w})

Ersetze zurück

= 3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} - \frac{1}{5 sin ^{5} ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )} - - \frac{2}{3 sin ^{3} ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )} - \frac{1}{sin ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )}

Vereinfache

3 \cdot 8 \cdot 2^{\frac{1}{2}} \cdot \frac{1}{9826} - \frac{1}{5 sin ^{5} ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )} - - \frac{2}{3 sin ^{3} ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )} - \frac{1}{sin ( arcsec ( \frac{2}{1 7 ^{\frac{1}{2}}} ( x + \frac{5}{2} ) ) )} : \frac{( 2 x + 5 ) ( 40 ( 2 x + 5 ) ^{2} ( x ^{2} + 5 x + 2 ) - 3 ( 2 x + 5 ) ^{4} - 240 ( x ^{2} + 5 x + 2 ) ^{2} )}{98260 \cdot 2 ^{\frac{1}{2}} ( x ^{2} + 5 x + 2 ) ^{\frac{5}{2}}}

= \frac{( 2 x + 5 ) ( 40 ( 2 x + 5 ) ^{2} ( x ^{2} + 5 x + 2 ) - 3 ( 2 x + 5 ) ^{4} - 240 ( x ^{2} + 5 x + 2 ) ^{2} )}{98260 \cdot 2 ^{\frac{1}{2}} ( x ^{2} + 5 x + 2 ) ^{\frac{5}{2}}}

Füge eine Konstante zur Lösung hinzu

= \frac{( 2 x + 5 ) ( 40 ( 2 x + 5 ) ^{2} ( x ^{2} + 5 x + 2 ) - 3 ( 2 x + 5 ) ^{4} - 240 ( x ^{2} + 5 x + 2 ) ^{2} )}{98260 \cdot 2 ^{\frac{1}{2}} ( x ^{2} + 5 x + 2 ) ^{\frac{5}{2}}} + C

\int x (1 - 5 x^{2})^{6} d x

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