integral of (3x+5)/((2x^2+x+1)^2)

Lösung

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

64 (\frac{17}{224 7} arctan (\frac{1}{7} (4 x + 1)) + \frac{17}{448 7} sin (2 arctan (\frac{1}{7} (4 x + 1))) - \frac{3}{32 ( 16 x ^{2} + 8 x + 8 )}) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

2 x^{2} + x + 1 : 2 (x + \frac{1}{4})^{2} + \frac{7}{8}

= \int \frac{3 x + 5}{( 2 ( x + \frac{1}{4} ) ^{2} + \frac{7}{8} ) ^{2}} d x

Wende U-Substitution an

= \int \frac{64 ( 3 ( u - \frac{1}{4} ) + 5 )}{( 16 u ^{2} + 7 ) ^{2}} d u

Entferne die Konstante:

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

= 64 \cdot \int \frac{3 ( u - \frac{1}{4} ) + 5}{( 16 u ^{2} + 7 ) ^{2}} d u

Multipliziere aus

\frac{3 ( u - \frac{1}{4} ) + 5}{( 16 u ^{2} + 7 ) ^{2}} : \frac{3 ( u - \frac{1}{4} )}{( 16 u ^{2} + 7 ) ^{2}} + \frac{5}{( 16 u ^{2} + 7 ) ^{2}}

Wende die Summenregel an:

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

= 64 (\int \frac{3 ( u - \frac{1}{4} )}{( 16 u ^{2} + 7 ) ^{2}} d u + \int \frac{5}{( 16 u ^{2} + 7 ) ^{2}} d u)

\int \frac{3 ( u - \frac{1}{4} )}{( 16 u ^{2} + 7 ) ^{2}} d u = 3 (- \frac{1}{32 ( 16 u ^{2} + 7 )} - \frac{1}{448 7} (2 arctan (\frac{4}{7} u) + sin (2 arctan (\frac{4}{7} u))))

\int \frac{5}{( 16 u ^{2} + 7 ) ^{2}} d u = \frac{5}{112 7} (2 arctan (\frac{4}{7} u) + sin (2 arctan (\frac{4}{7} u)))

= 64 (3 (- \frac{1}{32 ( 16 u ^{2} + 7 )} - \frac{1}{448 7} (2 arctan (\frac{4}{7} u) + sin (2 arctan (\frac{4}{7} u)))) + \frac{5}{112 7} (2 arctan (\frac{4}{7} u) + sin (2 arctan (\frac{4}{7} u))))

Setze in

u = x + \frac{1}{4}

ein

= 64 3 - \frac{1}{32 ( 16 ( x + \frac{1}{4} ) ^{2} + 7 )} - \frac{1}{448 7} (2 arctan (\frac{4}{7} (x + \frac{1}{4})) + sin (2 arctan (\frac{4}{7} (x + \frac{1}{4})))) + \frac{5}{112 7} (2 arctan (\frac{4}{7} (x + \frac{1}{4})) + sin (2 arctan (\frac{4}{7} (x + \frac{1}{4}))))

Vereinfache

64 3 - \frac{1}{32 ( 16 ( x + \frac{1}{4} ) ^{2} + 7 )} - \frac{1}{448 7} (2 arctan (\frac{4}{7} (x + \frac{1}{4})) + sin (2 arctan (\frac{4}{7} (x + \frac{1}{4})))) + \frac{5}{112 7} (2 arctan (\frac{4}{7} (x + \frac{1}{4})) + sin (2 arctan (\frac{4}{7} (x + \frac{1}{4})))) : 64 (\frac{17}{224 7} arctan (\frac{1}{7} (4 x + 1)) + \frac{17}{448 7} sin (2 arctan (\frac{1}{7} (4 x + 1))) - \frac{3}{32 ( 16 x ^{2} + 8 x + 8 )})

= 64 (\frac{17}{224 7} arctan (\frac{1}{7} (4 x + 1)) + \frac{17}{448 7} sin (2 arctan (\frac{1}{7} (4 x + 1))) - \frac{3}{32 ( 16 x ^{2} + 8 x + 8 )})

Füge eine Konstante zur Lösung hinzu

= 64 (\frac{17}{224 7} arctan (\frac{1}{7} (4 x + 1)) + \frac{17}{448 7} sin (2 arctan (\frac{1}{7} (4 x + 1))) - \frac{3}{32 ( 16 x ^{2} + 8 x + 8 )}) + C

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