Correction exercice I

1. Calculons les intégrales suivantes :
a) \(\int {\frac{{dx}}{{2{x^2} + 8x + 20}}} \) \( = \frac{1}{2}\) \(\int {\frac{{dx}}{{\left[ {{{\left( {x + 2} \right)}^2} + 6} \right]}}} \)
Posons \(t = x + 2\), on a \(dt = dx\)
\(\frac{1}{2}\int {\frac{{dx}}{{\left[ {{{\left( {x + 2} \right)}^2} + 6} \right]}}} \) \( = \frac{1}{2}\frac{1}{{\sqrt 6 }}\) \(\arctan \frac{{x + 2}}{{\sqrt 6 }} + cte\)

Correction exercice II

Calculons les intégrales suivantes :
\(\int {\frac{{x + 3}}{{{x^2} - 2x - 5}}} dx\) \( = \) \(\int {\frac{{\frac{1}{2}(2x - 2) + \left( {3 + \frac{1}{2}2} \right)}}{{{x^2} - 2x - 5}}dx} \) \( = \frac{1}{2}\) \(\int {\frac{{(2x - 2)}}{{{x^2} - 2x - 5}}dx + } \) \(4\int {\frac{{dx}}{{{x^2} - 2x - 5}}} \). Ainsi
\(\frac{1}{2}\int {\frac{{(2x - 2)}}{{{x^2} - 2x - 5}}dx + } \) \(4\int {\frac{{dx}}{{{x^2} - 2x - 5}}} \) \( = \frac{1}{2}Log\left| {{x^2} - 2x - 5} \right|\) \( + \frac{2}{{\sqrt 6 }}\) \(Log\left| {\frac{{\sqrt 6 - (x - 1)}}{{\sqrt 6 + (x - 1)}}} \right|\) \( + cte\)

Correction exercice III

Calculons les intégrales suivantes :
\(I = \) \(\int {\frac{{dx}}{{\sqrt {{x^2} + 4x + 10} }}} \) \( = \int {\frac{{dx}}{{\sqrt {{{\left( {x + 2} \right)}^2} + 6} }}} \) \( = Log\) \(\left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right|\) + cte

Correction exercice IV

Calculons les intégrales suivantes :
\(I = \) \(\int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} \) \( = \int {\frac{{\frac{5}{2}(2x + 4) - 7}}{{\sqrt {{x^2} + 4x + 10} }}} \) \( = \frac{5}{2}\int {\frac{{(2x + 4)}}{{\sqrt {{x^2} + 4x + 10} }}} \) \( - 7\int {\frac{{dx}}{{\sqrt {{{\left( {x + 2} \right)}^2} + 6} }}} \) \( = 5\sqrt {{x^2} + 4x + 10} \) \( + Log\) \(\left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right|\) \( + cte\)

Correction exercice V

Calculons les intégrales suivantes :
a) \(I = \) \(\int {\sqrt {{a^2} - {x^2}} dx} \) \( = \) \(\int {\frac{{{a^2} - {x^2}}}{{\sqrt {{a^2} - {x^2}} }}dx} \) \( = {a^2}\) \(\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} \) \( - \int {\frac{{{x^2}dx}}{{\sqrt {{a^2} - {x^2}} }}} \)
\(\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} \) \( = \arcsin \frac{x}{a}\)
\(\int {\frac{{{x^2}dx}}{{\sqrt {{a^2} - {x^2}} }}} \) s’intègre par parties en prenant \(u = x\) \( \Rightarrow du = dx\) et \(v = \frac{x}{{\sqrt {{a^2} - {x^2}} }}\) \( \Rightarrow dv = \) \( - \sqrt {{a^2} - {x^2}} \)
\(I = \) \(\int {\sqrt {{a^2} - {x^2}} dx} \) \( = {a^2}\) \(\arcsin \frac{x}{a} + \) \(x\sqrt {{a^2} - {x^2}} \) \( - \int {\sqrt {{a^2} - {x^2}} dx} \)
Ainsi \(2\int {\sqrt {{a^2} - {x^2}} dx} \) \( = {a^2}\arcsin \frac{x}{a}\) \( + x\sqrt {{a^2} - {x^2}} \)
\(\int {\sqrt {{a^2} - {x^2}} dx} \) \( = \frac{{{a^2}}}{2}\arcsin \frac{x}{a}\) \( + \frac{x}{2}\sqrt {{a^2} - {x^2}} \) \( + cte\)
b) Soit à calculer la primitive de \(I = \) \(\int {\frac{{dx}}{{(x + 1)\sqrt {{x^2} + 4} }}} \)
Posons \(x + 1 = \) \(\frac{1}{u} \Rightarrow dx = \) \( - \frac{1}{{{u^2}}}du\)
Ainsi \(I = \) \(\int {\frac{{ - \frac{1}{{{u^2}}}du}}{{\frac{1}{u}\sqrt {{{\left( {\frac{1}{u} - 1} \right)}^2} + 4} }}} \) \( = - \frac{1}{{\sqrt 5 }}\) \(\int {\frac{{du}}{{\sqrt {{u^2} - \frac{2}{5}u + \frac{1}{5}} }}} \) \( = - \frac{1}{{\sqrt 5 }}\) \(\int {\frac{{du}}{{\sqrt {{{\left( {u - \frac{1}{5}} \right)}^2} + \frac{4}{{25}}} }}} \)
Nous obtenons finalement
\(I = - \frac{1}{{\sqrt 5 }}Log\) \(\left| {\frac{1}{2}(y + \sqrt {{y^2} + 4} )} \right|\) \( + cte\)
Avec \(y = \frac{5}{{x + 1}} - 1\)