答案来自数理数一数二团,希望可以帮到你!
∫[(x²sinx)/(x²+1)+1]dx的Mathematica程序为:
[Integral]((x^2*Sin[x])/(x^2+1)+1)[DifferentialD]x
结果为:-(1/(4E))((-1+E^2)CosIntegral[I-x]+(-1+E^2)CosIntegral[I+x]+I(4IE(x-Cos[x])+(1+E^2)SinIntegral[I-x]+(1+E^2)SinIntegral[I+x]))C
手算的具体过程:
∫[(x²sinx)/(x²+1)+1]dx=∫x²sinx/(x²+1)dx+∫dx
=∫[(x²+1)sinx-sinx]/(x²+1)dx+x
=∫sinxdx-∫sinx/(x²+1)dx+x
=-cosx-∫sinxdarctanx+x
接着用分部积分法去做……,这只是思路,呵呵!