Question

高等数学不定积分问题 求解 详细一点谢谢

第三题

Answer 1

x^4/(1-x^4)=-1+1/(1-x^4)=-1+0.5/(1-x²)+0.5/(1+x²)=-1+0.25/(1-x)+0.25/(1+x)+0.5/(1+x²)直接积分,得原函数=-x-0.25ln|1-x|+0.25ln|1+x|+0.5arctanx+c

Answer 2

∫ f(x) dx = ln[x-√(x^2+1) ]
f(x)
={1/[x-√(x^2+1) ] } . [ 1 - x/√(x^2+1) ]
= -1/√(x^2+1)
∫xf'(x) dx
=∫xdf(x)
=xf(x) -∫ f(x) dx
= -x/√(x^2+1) - ∫ f(x) dx
= -x/√(x^2+1) - ln[x-√(x^2+1) ] + C

Answer 3