求不定积分 ∫[(log₂x)²]dx
令log₂x=u,则x=2^u,dx=d(2^u)=(2^u)ln2du,于是
原式=∫u²d(2^u)=u²(2^u)-2∫u(2^u)du
=u²(2^u)-(2/ln2)∫ud(2^u)
=u²(2^u)-(2/ln2)[u(2^u)-∫(2^u)du]
=u²(2^u)-(2/ln2)[u(2^u)-2^u/ln2]+C
=x(log₂x)²-(2/ln2)[x(log₂x)-x/ln2]+C