Change display of integration example

analyse/main.tex

@@ -274,12 +274,15 @@
             \paragraph{Exemple}
 
             \begin{align*}
-                &\int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{, en posant }u=e^x \\\\
-                &= \int_{e^0}^{e^{\ln{\sqrt{3}}}} \frac{1}{u + \frac{1}{u}}\,\frac{\mathrm{d}u}{u}
-                = \int_1^{\sqrt{3}} \frac{1}{u^2 + 1}\,\mathrm{d}u \\
-                &= [\arctan{u}]_1^{\sqrt{3}}
-                = \arctan{\sqrt{3}} - \arctan{1} \\
-                &= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}
+                \int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{,\quad on pose }u=e^x
+            \end{align*}
+
+            \begin{align*}
+                &= \int_{e^0}^{e^{\ln{\sqrt{3}}}} \frac{1}{u + \frac{1}{u}}\,\frac{\mathrm{d}u}{u} \\\\
+                &= \int_1^{\sqrt{3}} \frac{1}{u^2 + 1}\,\mathrm{d}u
+                = [\arctan{u}]_1^{\sqrt{3}}
+                = \arctan{\sqrt{3}} - \arctan{1}
+                = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}
             \end{align*}
 
 \section{Equations différentielles}