Question

3到6题求不定积分

Answer 1

=∫tan√(1+x²)d√(1+x²)
=-ln|cos√(1+x²)|+C

追答

=∫1/(1+2lnx)dlnx
=(1/2)ln(1+2lnx)+C

=1/2∫x-xcos2xdx
=x²/4-1/4∫xdsin2x
=x²/4-xsin2x/4+1/4∫sin2xdx
=x²/4-xsin2x/4-cos2x/8+C

追问

3到6

麻烦了

追答

=xarctan2x-∫2x/(1+4x²)dx
=xarctan2x-(1/4)ln(1+4x²)+C