Question

求不定积分！

如图 求大神！

Answer 1

令t=arctan√x，则x=tan^2t，dx=2tantsec^2tdt
原式=∫tan^2t*t*2tantsec^2tdt
=2∫t*tan^3tsec^2tdt
=(1/2)*∫t*d(tan^4t)
=(t/2)*tan^4t-(1/2)*∫tan^4tdt
=(t/2)*tan^4t-(1/2)*∫tan^2t*(sec^2t-1)dt
=(t/2)*tan^4t-(1/2)*∫tan^2tsec^2tdt+(1/2)*∫tan^2tdt
=(t/2)*tan^4t-(1/2)*∫tan^2d(tant)+(1/2)*∫(sec^2t-1)dt
=(t/2)*tan^4t-(1/6)*tan^3t+(1/2)*tant-t/2+C
=(t/2)*(tan^4t-1)+(1/6)*tant*(3-tan^2t)+C
=(1/2)*arctan√x*(x^2-1)+(1/6)*√x*(3-x)+C，其中C是任意常数