Calculer :
I 1 = ∫ 0 1 x 2 ln ( x 2 + 1 ) dx I_1 = int from 0 to 1 {x^2 ln (x^2 +1 ) dx }
On pose { u ( x ) = ln ( x 2 + 1 ) v ' ( x ) = x 2 alignl left lbrace matrix{u(x)=ln(x^2 + 1)##v'(x)=x^2} right none d'où { u ' ( x ) = 2 x ( x 2 + 1 ) v ( x ) = x 3 3 alignl left lbrace matrix{u'(x)= {alignc {2 x} over (x^2 + 1)}## v(x)={alignc x^3 over 3}} right none
On obtient :
I 1 = [ x 3 3 ln ( x 2 + 1 ) ] 0 1 − 2 3 ∫ 0 1 x 4 x 2 + 1 dx I_1 = left [ x^3 over 3 ln (x^2 +1 )right ]_0^1 - 2 over 3 int from 0 to 1 {x^4 over {x^2 +1 } dx }
On effectue la division de x 4 x^4 par x 2 + 1 x^2 +1 .
On obtient : x 4 = ( x 2 + 1 ) ( x 2 − 1 ) + 1 x^4 = (x^2+1)(x^2-1) +1 . Donc
I 1 = ln 2 3 − 2 3 ∫ 0 1 ( x 2 − 1 + 1 x 2 + 1 ) dx I_1 = {ln 2} over 3 - 2 over 3 int from 0 to 1 left (x^2 - 1 + 1 over {x^2 +1 }right ) dx
Et donc :
I 1 = ln 2 3 − 2 3 [ x 2 3 − x + arctan x ] 0 1 = ln 2 3 + 4 9 − π 6 I_1 = {ln 2} over 3 - 2 over 3 left[ x^2 over 3 - x + arctan x right ]_0^1 = {ln 2} over 3 + 4 over 9 - %pi over 6