En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 2Intégration en mathématiques/Exercices/Intégrales 2 », n'a pu être restituée correctement ci-dessus.
∫
−
2
2
x
+
2
d
x
{\displaystyle \int _{-2}^{2}{\sqrt {x+2}}\,\mathrm {d} x}
Solution
∫
−
2
2
x
+
2
d
x
=
2
3
[
(
x
+
2
)
3
/
2
]
−
2
2
=
16
3
{\displaystyle \int _{-2}^{2}{\sqrt {x+2}}\,\mathrm {d} x={\frac {2}{3}}\left[(x+2)^{3/2}\right]_{-2}^{2}={\frac {16}{3}}}
∫
−
4
0
x
x
+
4
d
x
{\displaystyle \int _{-4}^{0}x{\sqrt {x+4}}\,\mathrm {d} x}
Solution
∫
−
4
0
x
x
+
4
d
x
=
2
∫
0
2
(
y
2
−
4
)
y
2
d
y
=
2
[
y
5
5
−
4
y
3
3
]
0
2
=
−
128
15
{\displaystyle \int _{-4}^{0}x{\sqrt {x+4}}\,\mathrm {d} x=2\int _{0}^{2}\left(y^{2}-4\right)y^{2}\,\mathrm {d} y=2\left[{\frac {y^{5}}{5}}-{\frac {4y^{3}}{3}}\right]_{0}^{2}=-{\frac {128}{15}}}
∫
2
4
d
x
x
2
−
1
{\displaystyle \int _{2}^{4}{\frac {\mathrm {d} x}{x^{2}-1}}}
Solution
∫
2
4
d
x
x
2
−
1
=
1
2
∫
2
4
(
1
x
−
1
−
1
x
+
1
)
d
x
=
1
2
[
ln
(
x
−
1
x
+
1
)
]
2
4
=
1
2
ln
(
9
5
)
=
ln
3
−
ln
5
2
{\displaystyle \int _{2}^{4}{\frac {\mathrm {d} x}{x^{2}-1}}={\frac {1}{2}}\int _{2}^{4}\left({\frac {1}{x-1}}-{\frac {1}{x+1}}\right)\,\mathrm {d} x={\frac {1}{2}}\left[\ln \left({\frac {x-1}{x+1}}\right)\right]_{2}^{4}={\frac {1}{2}}\ln \left({\frac {9}{5}}\right)=\ln 3-{\frac {\ln 5}{2}}}
∫
0
1
d
t
4
−
t
2
{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{4-t^{2}}}}
Solution
∫
0
1
d
t
4
−
t
2
=
1
4
∫
0
1
(
1
t
+
2
−
1
t
−
2
)
d
t
=
1
4
[
ln
(
t
+
2
|
t
−
2
|
)
]
0
1
=
ln
3
4
{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{4-t^{2}}}={\frac {1}{4}}\int _{0}^{1}\left({\frac {1}{t+2}}-{\frac {1}{t-2}}\right)\,\mathrm {d} t={\frac {1}{4}}\left[\ln \left({\frac {t+2}{|t-2|}}\right)\right]_{0}^{1}={\frac {\ln 3}{4}}}
∫
0
1
x
2
+
2
x
−
1
x
+
1
d
x
{\displaystyle \int _{0}^{1}{\frac {x^{2}+2x-1}{x+1}}\,\mathrm {d} x}
Solution
∫
0
1
x
2
+
2
x
−
1
x
+
1
d
x
=
∫
0
1
(
x
+
1
−
2
x
+
1
)
d
x
=
[
x
2
2
+
x
−
2
ln
(
x
+
1
)
]
0
1
=
3
2
−
2
ln
2
{\displaystyle \int _{0}^{1}{\frac {x^{2}+2x-1}{x+1}}\,\mathrm {d} x=\int _{0}^{1}\left(x+1-{\frac {2}{x+1}}\right)\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}+x-2\ln(x+1)\right]_{0}^{1}={\frac {3}{2}}-2\ln 2}
∫
0
1
t
3
−
4
t
2
+
2
t
+
1
t
+
1
d
t
{\displaystyle \int _{0}^{1}{\frac {t^{3}-4t^{2}+2t+1}{t+1}}\,\mathrm {d} t}
Solution
∫
0
1
t
3
−
4
t
2
+
2
t
+
1
t
+
1
d
t
=
∫
0
1
(
t
2
−
5
t
+
7
−
6
t
+
1
)
d
t
=
[
t
3
3
−
5
t
2
2
+
7
t
−
6
ln
(
t
+
1
)
]
0
1
=
29
6
−
6
ln
2
{\displaystyle \int _{0}^{1}{\frac {t^{3}-4t^{2}+2t+1}{t+1}}\,\mathrm {d} t=\int _{0}^{1}\left(t^{2}-5t+7-{\frac {6}{t+1}}\right)\,\mathrm {d} t=\left[{\frac {t^{3}}{3}}-{\frac {5t^{2}}{2}}+7t-6\ln(t+1)\right]_{0}^{1}={\frac {29}{6}}-6\ln 2}
∫
1
3
d
t
t
−
1
+
t
+
1
{\displaystyle \int _{1}^{3}{\frac {\mathrm {d} t}{{\sqrt {t-1}}+{\sqrt {t+1}}}}}
Solution
En multipliant par l'expression conjuguée du dénominateur :
∫
1
3
d
t
t
−
1
+
t
+
1
=
∫
1
3
t
−
1
−
t
+
1
(
t
−
1
)
−
(
t
+
1
)
d
t
=
1
2
∫
1
3
(
t
+
1
−
t
−
1
)
d
t
=
1
3
[
(
t
+
1
)
3
/
2
−
(
t
−
1
)
3
/
2
]
1
3
=
4
3
(
2
−
2
)
{\displaystyle \int _{1}^{3}{\frac {\mathrm {d} t}{{\sqrt {t-1}}+{\sqrt {t+1}}}}=\int _{1}^{3}{\frac {{\sqrt {t-1}}-{\sqrt {t+1}}}{(t-1)-(t+1)}}\,\mathrm {d} t={\frac {1}{2}}\int _{1}^{3}({\sqrt {t+1}}-{\sqrt {t-1}})\,\mathrm {d} t={\frac {1}{3}}\left[(t+1)^{3/2}-(t-1)^{3/2}\right]_{1}^{3}={\frac {4}{3}}\left(2-{\sqrt {2}}\right)}
∫
0
2
x
d
x
2
−
x
+
2
+
x
{\displaystyle \int _{0}^{2}{\frac {x\,\mathrm {d} x}{{\sqrt {2-x}}+{\sqrt {2+x}}}}}
Solution
∫
0
2
x
d
x
2
−
x
+
2
+
x
=
1
2
∫
0
2
(
2
+
x
−
2
−
x
)
d
x
=
1
3
[
(
2
+
x
)
3
/
2
+
(
2
−
x
)
3
/
2
]
0
2
=
4
3
(
2
−
2
)
{\displaystyle \int _{0}^{2}{\frac {x\,\mathrm {d} x}{{\sqrt {2-x}}+{\sqrt {2+x}}}}={\frac {1}{2}}\int _{0}^{2}\left({\sqrt {2+x}}-{\sqrt {2-x}}\right)\,\mathrm {d} x={\frac {1}{3}}\left[(2+x)^{3/2}+(2-x)^{3/2}\right]_{0}^{2}={\frac {4}{3}}(2-{\sqrt {2}})}
∫
2
3
x
+
x
−
1
x
+
1
x
2
−
1
d
x
{\displaystyle \int _{2}^{3}{\frac {x+{\sqrt {\frac {x-1}{x+1}}}}{x^{2}-1}}\,\mathrm {d} x}
Solution
∫
2
3
x
x
2
−
1
d
x
=
1
2
[
ln
(
x
2
−
1
)
]
2
3
=
1
2
ln
8
3
{\displaystyle \int _{2}^{3}{\frac {x}{x^{2}-1}}\,\mathrm {d} x={\frac {1}{2}}\left[\ln(x^{2}-1)\right]_{2}^{3}={\frac {1}{2}}\ln {\frac {8}{3}}}
∫
2
3
x
−
1
x
+
1
x
2
−
1
d
x
=
[
x
−
1
x
+
1
]
2
3
=
1
2
−
1
3
{\displaystyle \int _{2}^{3}{\frac {\sqrt {\frac {x-1}{x+1}}}{x^{2}-1}}\,\mathrm {d} x=\left[{\sqrt {\frac {x-1}{x+1}}}\right]_{2}^{3}={\frac {1}{\sqrt {2}}}-{\frac {1}{\sqrt {3}}}}
![{\displaystyle \int _{-2}^{2}{\sqrt {x+2}}\,\mathrm {d} x={\frac {2}{3}}\left[(x+2)^{3/2}\right]_{-2}^{2}={\frac {16}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df9f83b44db8ff1b7a4a9961f04e790e8327ea18)
![{\displaystyle \int _{-4}^{0}x{\sqrt {x+4}}\,\mathrm {d} x=2\int _{0}^{2}\left(y^{2}-4\right)y^{2}\,\mathrm {d} y=2\left[{\frac {y^{5}}{5}}-{\frac {4y^{3}}{3}}\right]_{0}^{2}=-{\frac {128}{15}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed968f31c4645a7583b2692935d497f5c9a6f091)
![{\displaystyle \int _{2}^{4}{\frac {\mathrm {d} x}{x^{2}-1}}={\frac {1}{2}}\int _{2}^{4}\left({\frac {1}{x-1}}-{\frac {1}{x+1}}\right)\,\mathrm {d} x={\frac {1}{2}}\left[\ln \left({\frac {x-1}{x+1}}\right)\right]_{2}^{4}={\frac {1}{2}}\ln \left({\frac {9}{5}}\right)=\ln 3-{\frac {\ln 5}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dafe9b6757942149b221a2b65f819ea6c7cb7f29)
![{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{4-t^{2}}}={\frac {1}{4}}\int _{0}^{1}\left({\frac {1}{t+2}}-{\frac {1}{t-2}}\right)\,\mathrm {d} t={\frac {1}{4}}\left[\ln \left({\frac {t+2}{|t-2|}}\right)\right]_{0}^{1}={\frac {\ln 3}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eae88454c75df87615b4352aa198497c08d2d5c2)
![{\displaystyle \int _{0}^{1}{\frac {x^{2}+2x-1}{x+1}}\,\mathrm {d} x=\int _{0}^{1}\left(x+1-{\frac {2}{x+1}}\right)\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}+x-2\ln(x+1)\right]_{0}^{1}={\frac {3}{2}}-2\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4b244ec62e30963060d44f26e8b88e3e2ba0099)
![{\displaystyle \int _{0}^{1}{\frac {t^{3}-4t^{2}+2t+1}{t+1}}\,\mathrm {d} t=\int _{0}^{1}\left(t^{2}-5t+7-{\frac {6}{t+1}}\right)\,\mathrm {d} t=\left[{\frac {t^{3}}{3}}-{\frac {5t^{2}}{2}}+7t-6\ln(t+1)\right]_{0}^{1}={\frac {29}{6}}-6\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa46da1f2f7d08a909a7e5c6406c65f963256c8)
![{\displaystyle \int _{1}^{3}{\frac {\mathrm {d} t}{{\sqrt {t-1}}+{\sqrt {t+1}}}}=\int _{1}^{3}{\frac {{\sqrt {t-1}}-{\sqrt {t+1}}}{(t-1)-(t+1)}}\,\mathrm {d} t={\frac {1}{2}}\int _{1}^{3}({\sqrt {t+1}}-{\sqrt {t-1}})\,\mathrm {d} t={\frac {1}{3}}\left[(t+1)^{3/2}-(t-1)^{3/2}\right]_{1}^{3}={\frac {4}{3}}\left(2-{\sqrt {2}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85e10e03d2eec79fc82134ae87d8668b8d58d2f0)
![{\displaystyle \int _{0}^{2}{\frac {x\,\mathrm {d} x}{{\sqrt {2-x}}+{\sqrt {2+x}}}}={\frac {1}{2}}\int _{0}^{2}\left({\sqrt {2+x}}-{\sqrt {2-x}}\right)\,\mathrm {d} x={\frac {1}{3}}\left[(2+x)^{3/2}+(2-x)^{3/2}\right]_{0}^{2}={\frac {4}{3}}(2-{\sqrt {2}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f7446a5de51b31ad0e6eda055ac2c073d859c7d)
![{\displaystyle \int _{2}^{3}{\frac {x}{x^{2}-1}}\,\mathrm {d} x={\frac {1}{2}}\left[\ln(x^{2}-1)\right]_{2}^{3}={\frac {1}{2}}\ln {\frac {8}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d87500d6e0524c7e9db893ed4fc3dc54afc39c54)
![{\displaystyle \int _{2}^{3}{\frac {\sqrt {\frac {x-1}{x+1}}}{x^{2}-1}}\,\mathrm {d} x=\left[{\sqrt {\frac {x-1}{x+1}}}\right]_{2}^{3}={\frac {1}{\sqrt {2}}}-{\frac {1}{\sqrt {3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d25e59baed55548440cf6903c704a5869e7115a)
