Calculer l'intégrale définie suivante :
I 4 = ∫ − 1 0 x − 1 x 2 + x + 1 dx I_4 = int from -1 to 0 {{x-1}over{x^2+x+1}} dx
On écrit x 2 + x + 1 x^2+x+1 sous forme canonique.
On obtient : x 2 + x + 1 = ( x + 1 2 ) 2 + 3 4 x^2+x+1 = left ( x + {1 over 2} right )^2 + 3 over 4
que l'on écrit sous la forme x 2 + x + 1 = 3 4 [ [ 2 3 ( x + 1 2 ) 2 ] + 1 ] x^2+x+1 = 3 over 4 left [ left [ 2 over sqrt 3 left ( x + 1 over 2 right )^2 right] +1 right ]
On pose t = 2 3 ( x + 1 2 ) ⇔ x = 3 2 t − 1 2 t = 2 over sqrt 3 left ( x + 1 over 2 right ) `dlrarrow` x={sqrt 3 } over 2 t - 1 over 2
On obtient : dx = 3 2 dt dx={sqrt 3 } over 2 dt
Valeurs aux bornes : x x varie de − 1 -1 à 0 0 , donc t t varie de − 1 3 -{1 over sqrt 3} à 1 3 1 over sqrt 3 .
On obtient : x − 1 x 2 + x + 1 dx = 3 2 t − 3 2 3 4 ( t 2 + 1 ) × 3 2 dt {x-1}over {x^2 + x + 1}dx ={{size 9 {{sqrt 3} over 2} t - size 9 {3 over 2 }}over {size 9 {3 over 4}(t^2+1)}} times{{sqrt 3} over 2 dt }
donc
I 4 = ∫ − 1 3 1 3 t − 3 ( t 2 + 1 ) dt = ∫ − 1 3 1 3 t ( t 2 + 1 ) dt − ∫ − 1 3 1 3 3 ( t 2 + 1 ) dt I_4 = int from size 7 {-{1 over {sqrt 3}}} to size 7 {1 over {sqrt 3}} {{t - sqrt 3}over(t^2+1)} dt = int from size 7 {-{1 over {sqrt 3}}} to size 7 {1 over {sqrt 3}} {t over(t^2+1)} dt - int from size 7 {-{1 over {sqrt 3}}} to size 7 {1 over {sqrt 3}} {{sqrt 3}over(t^2+1)} dt
Enfin :
I 4 = [ 1 2 ln ( t 2 + 1 ) ] − 1 3 1 3 − [ 3 arctan t ] − 1 3 1 3 = 0 − 3 ( π 6 + π 6 ) = − π 3 3 I_4 = left[ 1 over 2 ln(t^2 + 1 ) right ] _{size 7 {-{1 over {sqrt 3}}}}^{size 7 {{1 over {sqrt 3}}}} - left[ sqrt 3 arctan t right ] _{size 7 {-{1 over {sqrt 3}}}}^{size 7 {{1 over {sqrt 3}}}} =0- sqrt 3 left(%pi over 6 + %pi over 6 right) = -{{%pi sqrt 3} over 3}