integral of ((4-x))/(sqrt((x^2-2x-3)^3))

Lösung

\int \frac{( 4 - x )}{( x ^{2} - 2 x - 3 ) ^{3}} d x

\frac{1}{x ^{2} - 2 x - 3} - \frac{3 ( x - 1 )}{4 x ^{2} - 2 x - 3} + C

Schritte zur Lösung

\int \frac{4 - x}{( x ^{2} - 2 x - 3 ) ^{3}} d x

Vervollständige das Quadrat

x^{2} - 2 x - 3 : (x - 1)^{2} - 4

= \int \frac{4 - x}{( ( x - 1 ) ^{2} - 4 ) ^{3}} d x

Wende U-Substitution an

= \int \frac{- u + 3}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

Multipliziere aus

\frac{- u + 3}{( u ^{2} - 4 ) ^{\frac{3}{2}}} : - \frac{u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} + \frac{3}{( u ^{2} - 4 ) ^{\frac{3}{2}}}

= \int - \frac{u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} + \frac{3}{( u ^{2} - 4 ) ^{\frac{3}{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

= - \int \frac{u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u + \int \frac{3}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

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

\int \frac{3}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u = - \frac{3 u}{4 u ^{2} - 4}

= - (- \frac{1}{u ^{2} - 4}) - \frac{3 u}{4 u ^{2} - 4}

Setze in

u = x - 1

ein

= - - \frac{1}{( x - 1 ) ^{2} - 4} - \frac{3 ( x - 1 )}{4 ( x - 1 ) ^{2} - 4}

Vereinfache

- - \frac{1}{( x - 1 ) ^{2} - 4} - \frac{3 ( x - 1 )}{4 ( x - 1 ) ^{2} - 4} : \frac{1}{x ^{2} - 2 x - 3} - \frac{3 ( x - 1 )}{4 x ^{2} - 2 x - 3}

= \frac{1}{x ^{2} - 2 x - 3} - \frac{3 ( x - 1 )}{4 x ^{2} - 2 x - 3}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{x ^{2} - 2 x - 3} - \frac{3 ( x - 1 )}{4 x ^{2} - 2 x - 3} + C

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