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13. 令 u = x-2， 则 x = u+2, dx = du
I = ∫<0, 2> f(u)du = ∫<0, 1>uarctanudu + ∫<1, 2>udu
= (1/2)∫<0, 1>arctanudu^2 + [u^2/2]<1, 2>
= (1/2)[u^2arctanu]<0, 1> - (1/2)∫<0, 1>[u^2/(1+u^2)]du + 3/2
= (1/2)(π/4) - (1/2)∫<0, 1>[1-1/(1+u^2)]du + 3/2
= 3/2 + π/8 - 1/2 + (1/2)[arctanu]<0, 1>
= 1 + π/4
I = ∫<0, 2> f(u)du = ∫<0, 1>uarctanudu + ∫<1, 2>udu
= (1/2)∫<0, 1>arctanudu^2 + [u^2/2]<1, 2>
= (1/2)[u^2arctanu]<0, 1> - (1/2)∫<0, 1>[u^2/(1+u^2)]du + 3/2
= (1/2)(π/4) - (1/2)∫<0, 1>[1-1/(1+u^2)]du + 3/2
= 3/2 + π/8 - 1/2 + (1/2)[arctanu]<0, 1>
= 1 + π/4
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