11.22 1)
2) π Z +∞
0
x 4x3+ 1
2
dx=π Z +∞
0
x2
(4x3+ 1)2 dx=π lim
t→+∞
Z t
0
x2
(4x3+ 1)2 dx
=π lim
t→+∞
1 12
Z t
0
(4x3+ 1)−2·12x2dx= 12π lim
t→+∞
1
−1(4x3+ 1)−1
t
0
=
π 12 lim
t→+∞− 1 4x3+ 1
t
0
= 12π lim
t→+∞
− 1
4t3+ 1 + 1 4·03+ 1
=
π 12 lim
t→+∞
− 1
4t3+ 1 + 1
= 12π lim
t→+∞
− 1
4t3 + 1
= 12π (−0 + 1) = 12π
Analyse : intégrales Corrigé 11.22