${\displaystyle \int _{-2}^{2}{\sqrt {x+2}}\,\mathrm {d} x={\frac {2}{3}}\left[(x+2)^{3/2}\right]_{-2}^{2}={\frac {16}{3}}}$.

${\displaystyle \int _{-4}^{0}x{\sqrt {x+4}}\,\mathrm {d} x=2\int _{0}^{2}\left(y^{2}-4\right)y^{2}\,\mathrm {d} y=2\left[{\frac {y^{5}}{5}}-{\frac {4y^{3}}{3}}\right]_{0}^{2}=-{\frac {128}{15}}}$.

${\displaystyle \int _{2}^{4}{\frac {\mathrm {d} x}{x^{2}-1}}={\frac {1}{2}}\int _{2}^{4}\left({\frac {1}{x-1}}-{\frac {1}{x+1}}\right)\,\mathrm {d} x={\frac {1}{2}}\left[\ln \left({\frac {x-1}{x+1}}\right)\right]_{2}^{4}={\frac {1}{2}}\ln \left({\frac {9}{5}}\right)=\ln 3-{\frac {\ln 5}{2}}}$.

${\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{4-t^{2}}}={\frac {1}{4}}\int _{0}^{1}\left({\frac {1}{t+2}}-{\frac {1}{t-2}}\right)\,\mathrm {d} t={\frac {1}{4}}\left[\ln \left({\frac {t+2}{|t-2|}}\right)\right]_{0}^{1}={\frac {\ln 3}{4}}}$.

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

${\displaystyle \int _{0}^{1}{\frac {t^{3}-4t^{2}+2t+1}{t+1}}\,\mathrm {d} t=\int _{0}^{1}\left(t^{2}-5t+7-{\frac {6}{t+1}}\right)\,\mathrm {d} t=\left[{\frac {t^{3}}{3}}-{\frac {5t^{2}}{2}}+7t-6\ln(t+1)\right]_{0}^{1}={\frac {29}{6}}-6\ln 2}$.

En multipliant par l'expression conjuguée du dénominateur :

${\displaystyle \int _{1}^{3}{\frac {\mathrm {d} t}{{\sqrt {t-1}}+{\sqrt {t+1}}}}=\int _{1}^{3}{\frac {{\sqrt {t-1}}-{\sqrt {t+1}}}{(t-1)-(t+1)}}\,\mathrm {d} t={\frac {1}{2}}\int _{1}^{3}({\sqrt {t+1}}-{\sqrt {t-1}})\,\mathrm {d} t={\frac {1}{3}}\left[(t+1)^{3/2}-(t-1)^{3/2}\right]_{1}^{3}={\frac {4}{3}}\left(2-{\sqrt {2}}\right)}$

${\displaystyle \int _{0}^{2}{\frac {x\,\mathrm {d} x}{{\sqrt {2-x}}+{\sqrt {2+x}}}}={\frac {1}{2}}\int _{0}^{2}\left({\sqrt {2+x}}-{\sqrt {2-x}}\right)\,\mathrm {d} x={\frac {1}{3}}\left[(2+x)^{3/2}+(2-x)^{3/2}\right]_{0}^{2}={\frac {4}{3}}(2-{\sqrt {2}})}$.