diff --git a/analyse/main.tex b/analyse/main.tex
index a78f975..b4c8b0d 100644
--- a/analyse/main.tex
+++ b/analyse/main.tex
@@ -258,15 +258,19 @@
         \paragraph{Exemple}
 
             \begin{align*}
-                \int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{,\quad on pose }u=e^x
+                \int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{,\quad on pose }u&=e^x \\
+                u' &= e^x = \frac{\mathrm{d}u}{\mathrm{d}x} \\
+                \mathrm{d}x &= \frac{\mathrm{d}u}{e^x} = \frac{\mathrm{d}u}{u}
             \end{align*}
 
+            Cela nous donne~:
+
             \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}
+                \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*}
 
 \clearpage