已知f(x)的一个原函数为xsinx,求∫xf('x)dx

∵ ∫f(x)dx = xsinx + c [Given,已知]
∴ f(x) = sinx + xcosx [Derivative,求导]
∴ ∫xf'(x)dx = ∫xdf(x) [Completing differentiation,凑微分]
= xf(x) - ∫f(x)dx + c [Integration by parts,分部积分]
= x(sinx + xcosx) - xsinx + c [Substitution,代入]
= x²cosx + c [Simplification,化简]