Question

∫(1－1／x²)√x√xdx

Answer 1

∫(1-1/x²)√(x√x)dx
= ∫[x^(3/4) - x^(-5/4)]dx
= (4/7)x^(7/4) - (-4)x^(-1/4) + C
= 4(x^2+7)/[7x^(1/4)] + C

Answer 2

x*dx/(1+√1+x^2)
=x*(1-√1+x^2)*dx/[(1+√1+x^2)*(1-√1+x^2)]
=x*(1-√1+x^2)*dx/(-x^2)
=(-1/x)*dx+(1/x)(√1+x^2)*dx
(-1/x)*dx=-d(lnx)
(1/x)(√1+x^2)*dx
好久没接触积分了，如果没有现成公式的话，看长相应该是换元x=tany，尝试一下：
(cosy/siny)*(1/cosy)*(1/cosy^2)dy=1/(siny*cosy*cosy)dy
=siny/(siny*siny*cosy*cosy)dy
=-d(cosy)/(siny*siny*cosy*cosy)
再换元令z=cosy，得
-dy/[y^2*(1-y^2)]=-dy[1/y^2+1/(1-y^2)]=d(1/y)-dy/[(1+y)(1-y)]=d(1/y)-dy*[1/(1+y)+1/(1-y)]/2
=d(1/y)-d[ln(1+y)]/2+d[ln(1-y)]/2
总结：可解，第一个换元应该有更好的