Réponse : Bonjour,
1) Posons v(t)=1+t², alors v'(t)=2t. On remarque donc que:
[tex]\displaystyle \frac{t}{(1+t^{2})^{2}}=\frac{1}{2}\frac{v'(t)}{v(t)^{2}}[/tex]
Et on a:
[tex]\displaystyle \frac{v'(t)}{v(t)^{2}}=\left(-\frac{1}{v(t)}\right)'[/tex]
Donc une primitive de [tex]\displaystyle t \mapsto \frac{t}{(1+t^{2})^{2}}[/tex] sur [0;1] est:
[tex]\displaystyle -\frac{1}{2} \times \frac{1}{1+t^{2}}=-\frac{1}{2(1+t^{2})}\\[/tex]
2) On a:
[tex]\displaystyle K-L=\int_{0}^{1} \frac{dt}{1+t^{2}}-\int_{0}^{1} \frac{t^{2} dt}{(1+t^{2})^{2}}=\int_{0}^{1} \frac{1+t^{2}-t^{2}}{(1+t^{2})^{2}} \; dt=\int_{0}^{1} \frac{1}{(1+t^{2})^{2}} \; dt=I[/tex]
Donc I=K-L.
3) On a:
[tex]\displaystyle \int_{0}^{1} \frac{t^{2}}{(1+t^{2})^{2}} \; dt=\left[t \times - \frac{1}{2(1+t^{2})}\right]_{0}^{1}-\int_{0}^{1} 1 \times -\frac{1}{2(1+t^{2})} \; dt\\=-\frac{1}{4}+\frac{1}{2} \int_{0}^{1} \frac{1}{1+t^{2}} \; dt=-\frac{1}{4}+\frac{1}{2}K[/tex]
4) On a:
[tex]\displaystyle K=\int_{0}^{1} \frac{1}{1+t^{2}} \; dt=[\arctan(t)]_{0}^{1}=\arctan(1)-arctan(0)=\frac{\pi}{4}[/tex]
5) D'après la question 3:
[tex]\displaystyle L=-\frac{1}{4}+\frac{1}{2}K=-\frac{1}{4}+\frac{1}{2} \times \frac{\pi}{4}=-\frac{1}{4}+\frac{\pi}{8}=\frac{-2+\pi}{8}=\frac{1}{8}(\pi-2)[/tex]
D'après la question 2), I=K-L, donc:
[tex]\displaystyle I=K-L=\frac{\pi}{4}-\left(-\frac{1}{4}+\frac{\pi}{8}\right)=\frac{\pi}{4}+\frac{1}{4}-\frac{\pi}{8}=\frac{2\pi}{8}-\frac{\pi}{8}+\frac{1}{4}=\frac{\pi}{8}+\frac{1}{4}\\ I=\frac{\pi+2}{8}=\frac{1}{8}(\pi+2)[/tex]
