求不定积分:[∫(√x^2-9)⼀x]dx

2024-11-06 20:41:06
推荐回答(1个)
回答1:

=∫(√x^2-9)/2*x*x]d(x*x)=0.5*∫(√u-9)/u]du=
∫t*t/t*t 9]dt=t-3*arctan(t/3) c=(√x^2-9)-3*arctan((√x^2-9)/3) c