第2个回答 2011-10-30
求定积分[-e,1]∫ ln(1+x²)dx
解:为简化书写过程,先求不定积分:
∫ ln(1+x²)dx =xln(1+x²)-∫[2x²/(1+x²)]dx=xln(1+x²)-2∫[1-1/(1+x²)]dx=xln(1+x²)-2[x-arctanx]+C
故[-e,1]∫ ln(1+x²)dx =[xln(1+x²)-2(x-arctanx)+C]︱[-e,1]
=ln2-2(1-π/4)-[-eln(1+e²)-2(-e-arctan(-e)]
=ln2-2-π/2+eln(1+e²)-2e+2aarctane
=eln(1+e²)+ln2-2(e+1)-π/2+2aarctane