Calculer les intégrales suivantes.
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} .