数学不定积分,需要用分部积分法解

求不定积分∫(√x)arctan(√x)dx
令arctan(√x)=u,则√x=tanu,x=tan²u,dx=2tanusec²udu；
故原式=2∫utan²usec²udu=(2/3)∫ud(tan³u)=(2/3)[utan³u-∫tan³udu]
=(2/3)[utan³u-∫tanu(1-sec²u)du]=(2/3)[utan³u-∫tanudu+∫tanusec²udu]
=(2/3)[utan³u+∫d(cosu)/cosu+∫tanud(tanu)]
=(2/3)[utan³u+ln∣cosu∣+(1/2)tan²u]+C
=(2/3){x³/²arctan(√x)+ln[1/√(1+x)]+x/2}+C
=(2/3)[(x√x)arctan(√x)-(1/2)ln(1+x)+x/2]+C