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

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

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

${\displaystyle \left[\ln \left(x+\mathrm {e} ^{x}\right)\right]_{0}^{1}=\ln \left(1+\mathrm {e} \right)}$.

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

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

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

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

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

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

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

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

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