En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 7Intégration en mathématiques/Exercices/Intégrales 7 », n'a pu être restituée correctement ci-dessus.
∫
1
e
1
+
ln
t
t
d
t
{\displaystyle \int _{1}^{\mathrm {e} }{\frac {1+\ln t}{t}}\,\mathrm {d} t}
Solution
[
ln
t
+
ln
2
t
2
]
1
e
=
3
2
{\displaystyle \left[\ln t+{\frac {\ln ^{2}t}{2}}\right]_{1}^{\mathrm {e} }={\frac {3}{2}}}
∫
1
2
e
ln
t
t
d
t
{\displaystyle \int _{1}^{2}{\frac {\mathrm {e} ^{\ln t}}{t}}\,\mathrm {d} t}
Solution
∫
1
2
t
t
d
t
=
1
{\displaystyle \int _{1}^{2}{\frac {t}{t}}\,\mathrm {d} t=1}
∫
0
1
2
x
e
x
2
+
1
d
x
{\displaystyle \int _{0}^{1}2x\operatorname {e} ^{x^{2}+1}\,\mathrm {d} x}
Solution
[
e
x
2
+
1
]
0
1
=
e
2
−
e
{\displaystyle \left[\mathrm {e} ^{x^{2}+1}\right]_{0}^{1}=\mathrm {e} ^{2}-\mathrm {e} }
∫
0
1
1
+
e
x
x
+
e
x
d
x
{\displaystyle \int _{0}^{1}{\frac {1+\mathrm {e} ^{x}}{x+\mathrm {e} ^{x}}}\,\mathrm {d} x}
∫
0
1
1
+
e
x
(
x
+
e
x
)
3
/
2
d
x
{\displaystyle \int _{0}^{1}{\frac {1+\mathrm {e} ^{x}}{\left(x+\mathrm {e} ^{x}\right)^{3/2}}}\,\mathrm {d} x}
Solution
[
ln
(
x
+
e
x
)
]
0
1
=
ln
(
1
+
e
)
{\displaystyle \left[\ln \left(x+\mathrm {e} ^{x}\right)\right]_{0}^{1}=\ln \left(1+\mathrm {e} \right)}
∫
1
1
+
e
t
−
3
/
2
d
t
=
−
2
[
t
−
1
/
2
]
1
1
+
e
=
2
(
1
−
1
1
+
e
)
{\displaystyle \int _{1}^{1+\mathrm {e} }t^{-3/2}\,\mathrm {d} t=-2\left[t^{-1/2}\right]_{1}^{1+\mathrm {e} }=2\left(1-{\frac {1}{\sqrt {1+\mathrm {e} }}}\right)}
∫
1
2
x
+
1
x
2
+
x
ln
x
d
x
{\displaystyle \int _{1}^{2}{\frac {x+1}{x^{2}+x\ln x}}\,\mathrm {d} x}
Solution
∫
0
ln
2
e
t
+
1
e
t
+
t
d
t
=
[
ln
(
e
t
+
t
)
]
0
ln
2
=
ln
(
2
+
ln
2
)
{\displaystyle \int _{0}^{\ln 2}{\frac {\mathrm {e} ^{t}+1}{\mathrm {e} ^{t}+t}}\,\mathrm {d} t=\left[\ln \left(\mathrm {e} ^{t}+t\right)\right]_{0}^{\ln 2}=\ln \left(2+\ln 2\right)}
∫
0
1
2
e
2
x
d
x
(
1
+
e
2
x
)
ln
(
1
+
e
2
x
)
{\displaystyle \int _{0}^{1}{\frac {2\mathrm {e} ^{2x}\mathrm {d} x}{\left(1+\mathrm {e} ^{2x}\right)\ln \left(1+\mathrm {e} ^{2x}\right)}}}
Solution
∫
2
1
+
e
2
d
t
t
ln
t
=
∫
ln
2
ln
(
1
+
e
2
)
d
s
s
=
ln
(
ln
(
1
+
e
2
)
ln
2
)
{\displaystyle \int _{2}^{1+\mathrm {e} ^{2}}{\frac {\mathrm {d} t}{t\ln t}}=\int _{\ln 2}^{\ln \left(1+\mathrm {e} ^{2}\right)}{\frac {\mathrm {d} s}{s}}=\ln \left({\frac {\ln \left(1+\mathrm {e} ^{2}\right)}{\ln 2}}\right)}
∫
0
1
e
x
−
1
e
x
+
1
d
x
{\displaystyle \int _{0}^{1}{\frac {\mathrm {e} ^{x}-1}{\mathrm {e} ^{x}+1}}\,\mathrm {d} x}
Solution
∫
1
e
y
−
1
y
+
1
d
y
y
=
∫
1
e
(
2
y
+
1
−
1
y
)
d
y
=
[
2
ln
(
y
+
1
)
−
ln
y
]
1
e
=
−
1
+
2
ln
1
+
e
2
{\displaystyle \int _{1}^{\mathrm {e} }{\frac {y-1}{y+1}}\,{\frac {\mathrm {d} y}{y}}=\int _{1}^{\mathrm {e} }\left({\frac {2}{y+1}}-{\frac {1}{y}}\right)\,\mathrm {d} y=\left[2\ln(y+1)-\ln y\right]_{1}^{\mathrm {e} }=-1+2\ln {\frac {1+\mathrm {e} }{2}}}
∫
1
π
(
1
+
cos
x
)
d
x
(
x
+
sin
x
)
ln
(
x
+
sin
x
)
{\displaystyle \int _{1}^{\pi }{\frac {(1+\cos x)\,\mathrm {d} x}{(x+\sin x)\ln(x+\sin x)}}}
Solution
∫
1
+
sin
1
π
d
t
t
ln
t
=
∫
ln
(
1
+
sin
1
)
ln
π
d
s
s
=
ln
(
ln
π
ln
(
1
+
sin
1
)
)
{\displaystyle \int _{1+\sin 1}^{\pi }{\frac {\mathrm {d} t}{t\ln t}}=\int _{\ln \left(1+\sin 1\right)}^{\ln \pi }{\frac {\mathrm {d} s}{s}}=\ln \left({\frac {\ln \pi }{\ln \left(1+\sin 1\right)}}\right)}
∫
1
n
(
|
x
−
1
|
+
|
x
−
2
|
+
⋯
+
|
x
−
n
|
)
d
x
,
n
∈
N
∗
{\displaystyle \int _{1}^{n}\left(|x-1|+|x-2|+\cdots +|x-n|\right)\,\mathrm {d} x,\quad n\in \mathbb {N} ^{*}}
Solution
∑
k
=
1
n
−
1
∫
k
k
+
1
(
(
x
−
1
)
+
(
x
−
2
)
+
⋯
+
(
x
−
k
)
+
(
k
+
1
−
x
)
+
(
k
+
2
−
x
)
+
⋯
+
(
n
−
x
)
)
d
x
=
{\displaystyle \sum _{k=1}^{n-1}\int _{k}^{k+1}\left((x-1)+(x-2)+\dots +(x-k)+(k+1-x)+(k+2-x)+\dots +(n-x)\right)\,\mathrm {d} x=}
∑
k
=
1
n
−
1
∫
k
k
+
1
(
(
2
k
−
n
)
x
+
n
(
n
+
1
)
2
−
k
(
k
+
1
)
)
d
x
=
{\displaystyle \sum _{k=1}^{n-1}\int _{k}^{k+1}\left((2k-n)x+{\frac {n(n+1)}{2}}-k(k+1)\right)\,\mathrm {d} x=}
∑
k
=
1
n
−
1
(
(
2
k
−
n
)
2
k
+
1
2
+
n
(
n
+
1
)
2
−
k
(
k
+
1
)
)
=
{\displaystyle \sum _{k=1}^{n-1}\left((2k-n){\frac {2k+1}{2}}+{\frac {n(n+1)}{2}}-k(k+1)\right)=}
∑
k
=
1
n
−
1
(
k
2
−
n
k
+
n
2
2
)
=
(
n
−
1
)
n
(
2
n
−
1
)
6
−
n
(
n
−
1
)
n
2
+
n
2
2
(
n
−
1
)
=
(
n
−
1
)
n
(
2
n
−
1
)
6
{\displaystyle \sum _{k=1}^{n-1}\left(k^{2}-nk+{\frac {n^{2}}{2}}\right)={\frac {(n-1)n(2n-1)}{6}}-n{\frac {(n-1)n}{2}}+{\frac {n^{2}}{2}}(n-1)={\frac {(n-1)n(2n-1)}{6}}}
En effectuant des intégrations par parties, calculer les intégrales suivantes :
I
1
=
∫
0
2
(
2
x
+
1
)
e
x
d
x
,
I
2
=
∫
1
e
x
ln
2
x
d
x
,
I
3
=
∫
0
π
2
cos
x
e
x
d
x
,
I
4
=
∫
1
2
x
2
1
+
x
2
arctan
x
d
x
{\displaystyle I_{1}=\int _{0}^{2}(2x+1)\operatorname {e} ^{x}\,\mathrm {d} x,\quad I_{2}=\int _{1}^{\mathrm {e} }x\ln ^{2}x\,dx,\quad I_{3}=\int _{0}^{\frac {\pi }{2}}\cos x\operatorname {e} ^{x}\,\mathrm {d} x,\quad I_{4}=\int _{1}^{2}{\frac {x^{2}}{1+x^{2}}}\arctan x\,\mathrm {d} x}
![{\displaystyle \left[\ln t+{\frac {\ln ^{2}t}{2}}\right]_{1}^{\mathrm {e} }={\frac {3}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50ab80a4ab68d26f40c6c35ab1f2332343bce3d2)
![{\displaystyle \left[\mathrm {e} ^{x^{2}+1}\right]_{0}^{1}=\mathrm {e} ^{2}-\mathrm {e} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/874a3288e8854842064f3342be0894399f408b8f)
![{\displaystyle \left[\ln \left(x+\mathrm {e} ^{x}\right)\right]_{0}^{1}=\ln \left(1+\mathrm {e} \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d90af70d55c68d8401bb98c94a90243c3f201c06)
![{\displaystyle \int _{1}^{1+\mathrm {e} }t^{-3/2}\,\mathrm {d} t=-2\left[t^{-1/2}\right]_{1}^{1+\mathrm {e} }=2\left(1-{\frac {1}{\sqrt {1+\mathrm {e} }}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32eb65ade75d9c29d2219a0d1ed7b879af9c3f26)
![{\displaystyle \int _{0}^{\ln 2}{\frac {\mathrm {e} ^{t}+1}{\mathrm {e} ^{t}+t}}\,\mathrm {d} t=\left[\ln \left(\mathrm {e} ^{t}+t\right)\right]_{0}^{\ln 2}=\ln \left(2+\ln 2\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc556f2953ca2d683c9882338c6821655e58ff3f)
![{\displaystyle \int _{1}^{\mathrm {e} }{\frac {y-1}{y+1}}\,{\frac {\mathrm {d} y}{y}}=\int _{1}^{\mathrm {e} }\left({\frac {2}{y+1}}-{\frac {1}{y}}\right)\,\mathrm {d} y=\left[2\ln(y+1)-\ln y\right]_{1}^{\mathrm {e} }=-1+2\ln {\frac {1+\mathrm {e} }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7594ec2003bc858708b4e9a1477b2c6e592993d2)
![{\displaystyle \left[(2x+1)\operatorname {e} ^{x}\right]_{0}^{2}-\int _{0}^{2}2\operatorname {e} ^{x}\,\mathrm {d} x=\left[(2x-1)\operatorname {e} ^{x}\right]_{0}^{2}=3\mathrm {e} ^{2}+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17228581be5a80867e724a24b173502d5f243dc2)
![{\displaystyle \left[{\frac {x^{2}}{2}}\ln ^{2}x\right]_{1}^{\mathrm {e} }-\int _{1}^{\mathrm {e} }x\ln x\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}\ln ^{2}x-{\frac {x^{2}}{2}}\ln x\right]_{1}^{\mathrm {e} }+\int _{1}^{\mathrm {e} }{\frac {x}{2}}\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}\ln ^{2}x-{\frac {x^{2}}{2}}\ln x+{\frac {x^{2}}{4}}\right]_{1}^{\mathrm {e} }={\frac {\mathrm {e} ^{2}-1}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f4d07421a333cd45c5f334fa1b5e95e40a2317)
![{\displaystyle I_{3}=\left[\cos x\operatorname {e} ^{x}\right]_{0}^{\frac {\pi }{2}}+\int _{0}^{\frac {\pi }{2}}\sin x\operatorname {e} ^{x}\,\mathrm {d} x=\left[(\cos x+\sin x)\operatorname {e} ^{x}\right]_{0}^{\frac {\pi }{2}}-I_{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdda24f510c82948152434d4aaa6e70e89ff30cc)
![{\displaystyle I_{4}=\int _{1}^{2}\arctan x\,\mathrm {d} x-\int _{1}^{2}{\frac {\arctan x}{1+x^{2}}}\,\mathrm {d} x=\left[x\arctan x\right]_{1}^{2}-\int _{1}^{2}{\frac {x}{1+x^{2}}}\,\mathrm {d} x-{\frac {1}{2}}\left[(\arctan x)^{2}\right]_{1}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50519b8afbaf171c04cba4818e992d3e34156e5e)
![{\displaystyle =\left[x\arctan x\right]_{1}^{2}-{\frac {1}{2}}\left[\ln(1+x^{2})\right]_{1}^{2}-{\frac {1}{2}}\left[(\arctan x)^{2}\right]_{1}^{2}=2\arctan 2-{\frac {\pi }{4}}-{\frac {1}{2}}\ln {\frac {5}{2}}-{\frac {(\arctan 2)^{2}}{2}}+{\frac {\pi ^{2}}{32}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0202d123279bba156796af1e9733556c8c876db9)
