${\displaystyle \int _{0}^{1}x(x^{2}+1)^{5}\,\mathrm {d} x={\frac {1}{2}}\int _{1}^{2}y^{5}\,\mathrm {d} y={\frac {2^{6}-1^{6}}{12}}={\frac {21}{4}}}$.

${\displaystyle \int _{2}^{7}(x^{3}-7x+15)\,\mathrm {d} x={\frac {7^{4}-2^{4}}{4}}-7{\frac {7^{2}-2^{2}}{2}}+15(7-2)={\frac {2385}{4}}-{\frac {315}{2}}+75={\frac {2055}{4}}}$.

${\displaystyle \int _{0}^{1}(4x+2)(x^{2}+x-2)^{2}\,\mathrm {d} x=2\int _{-2}^{0}y^{2}\,\mathrm {d} y={\frac {16}{3}}}$.

${\displaystyle \int _{0}^{2}x^{2}(x^{3}-8)\,\mathrm {d} x={\frac {1}{3}}\int _{-8}^{0}y\,\mathrm {d} y=-{\frac {32}{3}}}$.

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

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

${\displaystyle \int _{-1}^{0}{\frac {2x+3}{(x^{2}+3x-1)^{2}}}\,\mathrm {d} x=\int _{-3}^{-1}{\frac {\mathrm {d} y}{y^{2}}}=1-{\frac {1}{3}}={\frac {2}{3}}}$.

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

${\displaystyle \int _{0}^{1}{\frac {3x+1}{\sqrt {x^{2}+1}}}\;\mathrm {d} x=\left[{\frac {3}{2}}\ln(x^{2}+1)+\arctan x\right]_{0}^{1}={\frac {3}{2}}\ln 2+{\frac {\pi }{4}}}$.