Intégration en mathématiques/Exercices/Intégrales 7

Calculer les intégrales suivantes.

Exercice 15-1

$\int _{1}^{\mathrm {e} }{\frac {1+\ln t}{t}}\,\mathrm {d} t$

Solution

$\left[\ln t+{\frac {\ln ^{2}t}{2}}\right]_{1}^{\mathrm {e} }={\frac {3}{2}}$.

 

Exercice 15-2

$\int _{1}^{2}{\frac {\mathrm {e} ^{\ln t}}{t}}\,\mathrm {d} t$

Solution

$\int _{1}^{2}{\frac {t}{t}}\,\mathrm {d} t=1$.

 

Exercice 15-3

$\int _{0}^{1}2x\operatorname {e} ^{x^{2}+1}\,\mathrm {d} x$

Solution

$\left[\mathrm {e} ^{x^{2}+1}\right]_{0}^{1}=\mathrm {e} ^{2}-\mathrm {e}$.

 

Exercice 15-4

1. $\int _{0}^{1}{\frac {1+\mathrm {e} ^{x}}{x+\mathrm {e} ^{x}}}\,\mathrm {d} x$
2. $\int _{0}^{1}{\frac {1+\mathrm {e} ^{x}}{\left(x+\mathrm {e} ^{x}\right)^{3/2}}}\,\mathrm {d} x$

Solution

1. $\left[\ln \left(x+\mathrm {e} ^{x}\right)\right]_{0}^{1}=\ln \left(1+\mathrm {e} \right)$.
2. $\int _{1}^{1+\mathrm {e} }t^{-3/2}\,\mathrm {d} t=-2\left[t^{-1/2}\right]_{1}^{1+\mathrm {e} }=2\left(1-{\frac {1}{\sqrt {1+\mathrm {e} }}}\right)$.

 

Exercice 15-5

$\int _{1}^{2}{\frac {x+1}{x^{2}+x\ln x}}\,\mathrm {d} x$

Solution

$\int _{0}^{\ln 2}{\frac {\mathrm {e} ^{t}+1}{\mathrm {e} ^{t}+t}}\,\mathrm {d} t=\left[\ln \left(\mathrm {e} ^{t}+t\right)\right]_{0}^{\ln 2}=\ln \left(2+\ln 2\right)$.

 

Exercice 15-6

$\int _{0}^{1}{\frac {2\mathrm {e} ^{2x}\mathrm {d} x}{\left(1+\mathrm {e} ^{2x}\right)\ln \left(1+\mathrm {e} ^{2x}\right)}}$

Solution

$\int _{2}^{1+\mathrm {e} ^{2}}{\frac {\mathrm {d} t}{t\ln t}}=\int _{\ln 2}^{\ln \left(1+\mathrm {e} ^{2}\right)}{\frac {\mathrm {d} s}{s}}=\ln \left({\frac {\ln \left(1+\mathrm {e} ^{2}\right)}{\ln 2}}\right)$.

 

Exercice 15-7

$\int _{0}^{1}{\frac {\mathrm {e} ^{x}-1}{\mathrm {e} ^{x}+1}}\,\mathrm {d} x$

Solution

$\int _{1}^{\mathrm {e} }{\frac {y-1}{y+1}}\,{\frac {\mathrm {d} y}{y}}=\int _{1}^{\mathrm {e} }\left({\frac {2}{y+1}}-{\frac {1}{y}}\right)\,\mathrm {d} y=\left[2\ln(y+1)-\ln y\right]_{1}^{\mathrm {e} }=-1+2\ln {\frac {1+\mathrm {e} }{2}}$.

 

Exercice 15-8

$\int _{1}^{\pi }{\frac {(1+\cos x)\,\mathrm {d} x}{(x+\sin x)\ln(x+\sin x)}}$

Solution

$\int _{1+\sin 1}^{\pi }{\frac {\mathrm {d} t}{t\ln t}}=\int _{\ln \left(1+\sin 1\right)}^{\ln \pi }{\frac {\mathrm {d} s}{s}}=\ln \left({\frac {\ln \pi }{\ln \left(1+\sin 1\right)}}\right)$.

 

Exercice 15-9

$\int _{1}^{n}\left(|x-1|+|x-2|+\cdots +|x-n|\right)\,\mathrm {d} x,\quad n\in \mathbb {N} ^{*}$

Solution

$\sum _{k=1}^{n-1}\int _{k}^{k+1}\left((x-1)+(x-2)+\dots +(x-k)+(k+1-x)+(k+2-x)+\dots +(n-x)\right)\,\mathrm {d} x=$

$\sum _{k=1}^{n-1}\int _{k}^{k+1}\left((2k-n)x+{\frac {n(n+1)}{2}}-k(k+1)\right)\,\mathrm {d} x=$

$\sum _{k=1}^{n-1}\left((2k-n){\frac {2k+1}{2}}+{\frac {n(n+1)}{2}}-k(k+1)\right)=$

$\sum _{k=1}^{n-1}\left(k^{2}-nk+{\frac {n^{2}}{2}}\right)={\frac {(n-1)n(2n-1)}{6}}-n{\frac {(n-1)n}{2}}+{\frac {n^{2}}{2}}(n-1)={\frac {(n-1)n(2n-1)}{6}}$.

 

Exercice 15-10

En effectuant des intégrations par parties, calculer les intégrales suivantes :

$I_{1}=\int _{0}^{2}(2x+1)\operatorname {e} ^{x}\,\mathrm {d} x,\quad I_{2}=\int _{1}^{\mathrm {e} }x\ln ^{2}x\,dx,\quad I_{3}=\int _{0}^{\frac {\pi }{2}}\cos x\operatorname {e} ^{x}\,\mathrm {d} x,\quad I_{4}=\int _{1}^{2}{\frac {x^{2}}{1+x^{2}}}\arctan x\,\mathrm {d} x$.

Solution

1. $\left[(2x+1)\operatorname {e} ^{x}\right]_{0}^{2}-\int _{0}^{2}2\operatorname {e} ^{x}\,\mathrm {d} x=\left[(2x-1)\operatorname {e} ^{x}\right]_{0}^{2}=3\mathrm {e} ^{2}+1$.
2. $\left[{\frac {x^{2}}{2}}\ln ^{2}x\right]_{1}^{\mathrm {e} }-\int _{1}^{\mathrm {e} }x\ln x\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}\ln ^{2}x-{\frac {x^{2}}{2}}\ln x\right]_{1}^{\mathrm {e} }+\int _{1}^{\mathrm {e} }{\frac {x}{2}}\,\mathrm {d} x=\left[{\frac {x^{2}}{2}}\ln ^{2}x-{\frac {x^{2}}{2}}\ln x+{\frac {x^{2}}{4}}\right]_{1}^{\mathrm {e} }={\frac {\mathrm {e} ^{2}-1}{4}}$.
3. $I_{3}=\left[\cos x\operatorname {e} ^{x}\right]_{0}^{\frac {\pi }{2}}+\int _{0}^{\frac {\pi }{2}}\sin x\operatorname {e} ^{x}\,\mathrm {d} x=\left[(\cos x+\sin x)\operatorname {e} ^{x}\right]_{0}^{\frac {\pi }{2}}-I_{3}$ donc $I_{3}={\frac {\operatorname {e} ^{\frac {\pi }{2}}-1}{2}}$.
4. $I_{4}=\int _{1}^{2}\arctan x\,\mathrm {d} x-\int _{1}^{2}{\frac {\arctan x}{1+x^{2}}}\,\mathrm {d} x=\left[x\arctan x\right]_{1}^{2}-\int _{1}^{2}{\frac {x}{1+x^{2}}}\,\mathrm {d} x-{\frac {1}{2}}\left[(\arctan x)^{2}\right]_{1}^{2}$
   $=\left[x\arctan x\right]_{1}^{2}-{\frac {1}{2}}\left[\ln(1+x^{2})\right]_{1}^{2}-{\frac {1}{2}}\left[(\arctan x)^{2}\right]_{1}^{2}=2\arctan 2-{\frac {\pi }{4}}-{\frac {1}{2}}\ln {\frac {5}{2}}-{\frac {(\arctan 2)^{2}}{2}}+{\frac {\pi ^{2}}{32}}$.