求不定积分∫(x^5）/√(1-x^2)dx补充：上标是1下标是0时答案是8/15

Question

求不定积分∫(x^5）/√(1-x^2) dx 补充：上标是1下标是0时答案是 8/15

Answer 1

设√(1-x²)=t,则当x=0时,t=1.当x=1时,t=0.x²=1-t²,xdx=-tdt
故 原式=∫(x²)²xdx/√(1-x²)
=∫(1-t²)²(-t)dt/t
=∫(1-t²)²dt
=∫(t^4-2t²+1)dt
=(x^5/5-2x³/3+t)│
=1/5-2/3+1
=8/15.