Question

一道不定积分，求助

∫[x^3/√(x^2+1)]dx

Answer 1

对于这种题目，应优先考虑第二换元法
【这题用第二换元法比较快】
令x=tant, t∈(-π/2,π/2)
原式=∫(tant)^3(sect)dt
=∫(tant)^2d(sect)
=∫[(sect)^2-1]d(sect)
=∫(sect)^2d(sect)-∫d(sect)
=(1/3)(sect)^3-sect+C
根据tant=x/1，作辅助三角形，得
sect=√(x^2+1)
所以，原式=(1/3)(x^2+1)^(3/2)-√(x^2+1)+C

Answer 2

∫[x^3/√(x^2+1)]dx
=∫[x^2/√(x^2+1)]xdx
=1/2∫[x^2/√(x^2+1)]d(x^2+1)
=1/2∫*1/2*x^2d√(x^2+1)
=1/4∫x^2d√(x^2+1)
=1/4(x^2*√(x^2+1)-∫√(x^2+1)dx^2)
=1/4(x^2*√(x^2+1)-∫√(x^2+1)d(x^2+1))
=1/4(x^2*√(x^2+1)-3/2√(x^2+1)^3)+C
=1/4x^2*√(x^2+1)-3/8√(x^2+1)^3+C