Exercice 14-04
Calculer les primitives suivantes:
  1. \(\displaystyle \int \frac{x+2}{\sqrt{3x+4}}\,dx\) (\(x\gt -\frac43\))
  2. \(\displaystyle \int \sqrt{\frac{x^2}{x^2-1}}\, dx\) (\(x\gt 1\))
  3. \(\displaystyle \int |x|\, dx\)
  4. \(\displaystyle \int\frac{\sqrt{x-4}}{x}\,dx\) (\(x\gt 4\))
  5. \(\displaystyle \int 2^{\log(x)}\,dx\) (\(x\gt 0\))

\(3x+4=t\)

\(x^2-1=t\)

\(\sqrt{x-4}=t\)

\(2=e^{\log(2)}\)

  1. En posant \(t=3x+4\), c'est-à-dire \(x=\varphi(t)=\frac{t-4}{3}\), avec \(\varphi'(t)=\frac13\), \[\begin{aligned} \int\frac{x+2}{\sqrt{3x+4}}\,dx &=\int\frac{\frac{t-4}{3}+2}{\sqrt{t}}\cdot\frac{1}{3}\,dt\\ &=\frac{1}{9}\int \frac{t+2}{\sqrt{t}}\,dt\\ &=\frac{1}{9}\int\bigl(\sqrt{t}+4\frac{1}{2\sqrt{t}}\bigr)\,dt\\ &=\frac{2}{27}t^{3/2}+\frac49 \sqrt{t}+C\\ &=\frac{2}{27}(3x+4)^{3/2}+\frac49 \sqrt{3x+4}+C\,. \end{aligned}\]
  2. En posant \(x^2-1=t\), c'est-à-dire \(x=\varphi(t)=\sqrt{t+1}\), avec \(\varphi'(t)=\frac{1}{2\sqrt{t+1}}\), \[\begin{aligned} \int \sqrt{\frac{x^2}{x^2-1}}\,dx &=\int\sqrt{\frac{t+1}{t}}\cdot\frac{1}{2\sqrt{t+1}}\,dt\\ &=\int\frac{1}{2\sqrt{t}}\,dt\\ &=\sqrt{t}+C=\sqrt{x^2-1}+C\,. \end{aligned}\]
  3. Si \(x\geqslant 0\), alors \[ \int |x|\,dx=\int x\,dx=\frac{x^2}{2}+C_+\,, \] et si \(x< 0\), alors \[ \int |x|\,dx=\int (-x)\,dx=-\frac{x^2}{2}+C_-\,. \] Pour que la primitive soit dérivable, on doit prendre \(C_+=C_-=C\) (car si \(C_+\neq C_-\), elle n'est pas continue en zéro). Ainsi, \[ \int|x|\,dx= \begin{cases} +\frac{x^2}{2}+C&\text{ si }x\geqslant 0\,,\\ -\frac{x^2}{2}+C&\text{ si }x <0\,. \end{cases} \]
  4. Avec \(t=\sqrt{x-4}\), \[\begin{aligned} \int\frac{\sqrt{x-4}}{x}\,dx &=\int \frac{t}{t^2+4}\cdot 2t\,dt\\ &=2\int\frac{t^2{\color{blue}+4-4}}{t^2+4}\,dt\\ &=2t-4\int\frac{1/2}{(t/2)^2+1}\,dt\\ &=2t-4 \arctan(t/2)+C\\ &=2\sqrt{x-4}-4\arctan(\sqrt{x-4}/2)+C\,. \end{aligned}\] (Sans utiliser le truc du ''\({\color{blue}+4-4}\)'', on fait simplement une décomposition en éléments simples de \(\frac{t^2}{t^2+1}\).)
  5. On écrit \[\begin{aligned} \int 2^{\log(x)}\,dx &=\int \bigl(e^{\log(2)}\bigr)^{\log(x)}\,dx\\ &=\int \bigl(e^{\log(x)}\bigr)^{\log(2)}\,dx\\ &=\int x^{\log(2)}\,dx\\ &=\frac{x^{\log(2)+1}}{\log(2)+1}+C\,. \end{aligned}\]