Correction exercice I
Calculons les intégrales suivantes
a) \(I = \) \(\int {\frac{{\sqrt x }}{{\sqrt[4]{{{x^3}}} + 1}}} dx\)
I peut aussi se mettre sous la forme \(\int {\frac{{{x^{\frac{1}{2}}}}}{{{x^{\frac{3}{4}}} + 1}}} dx\)
Le dénominateur des fractions \({\frac{3}{4}}\) et \({\frac{1}{2}}\) est 4 donc posons \(x = {t^4}\) \( \Rightarrow dx = \) \(4{t^3}dt\)
\(I = 4\) \(\int {\frac{{{t^5}}}{{{t^3} + 1}}} dt = \) \(4\int {{t^2}dt - \frac{4}{3}} \) \(\int {\frac{{dt}}{{t + 1}}} + \frac{4}{3}\) \(\int {\frac{{ - 2t + 1}}{{{t^2} - t + 1}}} dt\)
\(I = \frac{4}{3}\sqrt[4]{{{x^3}}}\) \( - \frac{4}{3}Log\left| {\sqrt[4]{{{x^3}}} + 1} \right|\) \( + cte\)
b) \(I = \) \(\int {\frac{{\sqrt {{x^3}} - \sqrt[3]{x}}}{{6\sqrt[4]{x}}}} dx\)
Le dénominateur des fractions \({^{\frac{3}{4}}}\), \({\frac{1}{3}}\) et \({\frac{1}{4}}\) est 12
\(I = \) \(\int {\frac{{\sqrt {{x^3}} - \sqrt[3]{x}}}{{6\sqrt[4]{x}}}} dx\) \( = \) \(\int {\frac{{{x^{\frac{3}{2}}} - {x^{\frac{1}{3}}}}}{{6{x^{\frac{1}{4}}}}}} dx\)
\(x = {t^{12}} \Rightarrow \) \(dx = 12{t^{11}}dx\)
\(I = \) \(\int {\frac{{{x^{\frac{3}{2}}} - {x^{\frac{1}{3}}}}}{{6{x^{\frac{1}{4}}}}}} dx\) \( = 2\int {\frac{{{t^{18}} - {t^4}}}{{{t^3}}}} \) \({t^{11}}dt = 2\) \(\int {({t^{26}} - {t^{12}})} dt\) \( = \frac{2}{{27}}{t^{27}} - \) \(\frac{2}{{13}}{t^{13}} + cte\)
\(I = \frac{2}{{27}}{x^{\frac{9}{4}}}\) \( - \frac{2}{{13}}{x^{\frac{{13}}{{12}}}} + cte\) \( = \frac{2}{{27}}\sqrt[4]{{{x^9}}}\) \( - \frac{2}{{13}}\sqrt[{12}]{{{x^{13}}}} + cte\)
c) \(I = - \frac{6}{{\sqrt[6]{x}}} + \) \(\frac{{12}}{{\sqrt[{12}]{x}}} + 2Log\left( x \right)\) \( - 24\) \(Log\left( {\sqrt[{12}]{x} + 1} \right)\) \( + cte\)
d) \(I = \frac{6}{5}\sqrt[6]{{{x^5}}}\) \( - \frac{3}{2}\sqrt[6]{{{x^4}}} + \) \(4\sqrt[6]{{{x^3}}} + \) \(6\sqrt[6]{x} - \) \(9Log\left( {\sqrt[6]{x} + 1} \right)\) \( + \frac{3}{2}Log\left( {\sqrt[6]{{{x^2}}} + 1} \right)\) \( + 3\arctan \left( {\sqrt[6]{x}} \right)\) \( + cte\)
e) Posons \({u^2} = \frac{{1 - x}}{{1 + x}}\) on trouvera
\(I = Log\frac{{u - 1}}{{u + 1}}\) \( - \frac{1}{{{u^2} - 1}} = \) \(Log\) \(\left| {\frac{{\sqrt {1 - x} + \sqrt {1 + x} }}{{\sqrt {1 - x} - \sqrt {x + 1} }}} \right|\) \( - \frac{{\sqrt {1 + x} }}{x} + cte\)
f ) \(I = \) \(\sqrt {3{x^2} - 7x - 6} \) \( + \frac{{11}}{{2\sqrt 3 }}Log\) \(\left| {x - \frac{7}{2} + \sqrt {{x^2} - \frac{7}{3}x - 2} } \right|\) \( + cte\)
