解答题
15.设y'=arctan(x-1)2，y(0)=0，求∫01y(x)dx．
【正确答案】∫01y(x)dx=xy(x)∫01-∫01xarctan(x-1)2dx
=y(1)-∫01(x-1)arctan(x-1)2d(x-1)-∫01arctan(x-1)2dx
=
∫01arctan(x-1)2d(x-1)2=
∫01arctantdt
=
(tarctant｜01-∫01
=y(1)-∫01(x-1)arctan(x-1)2d(x-1)-∫01arctan(x-1)2dx
=
∫01arctan(x-1)2d(x-1)2=∫01arctantdt=
(tarctant｜01-∫01【答案解析】