integral of 1/((x^2-6x+10)^{5/2)}

Lösung

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{\frac{5}{2}}} d x

\frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

Schritte zur Lösung

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{\frac{5}{2}}} d x

Vervollständige das Quadrat

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

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

Wende U-Substitution an

= \int \frac{1}{( u ^{2} + 1 ) ^{\frac{5}{2}}} d u

Trigonometrische Substitution anwenden

= \int cos^{3} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (1 - sin^{2} (v)) cos (v) d v

Wende U-Substitution an

= \int 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

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

\int 1 d w = w

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

= w - \frac{w ^{3}}{3}

Ersetze zurück

= sin (arctan (x - 3)) - \frac{sin ^{3} ( arctan ( x - 3 ) )}{3}

Vereinfache

sin (arctan (x - 3)) - \frac{sin ^{3} ( arctan ( x - 3 ) )}{3} : \frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}}

= \frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}}

Füge eine Konstante zur Lösung hinzu

= \frac{x - 3}{( x ^{2} - 6 x + 10 ) ^{\frac{1}{2}}} - \frac{( x - 3 ) ^{3}}{3 ( x ^{2} - 6 x + 10 ) ^{\frac{3}{2}}} + C

\int \frac{cos ( 2 x )}{sin ^{2} ( x )} d x

\int \frac{sin ( x )}{sin ^{3} ( x ) + cos ^{3} ( x )} d x

\int \frac{1}{( x ^{2} + 2 x + 5 ) ^{2}} d x

\int \frac{1}{3 + 2 cos ( x )} d x

\int sin (2 x) cos (2 x) d x