Question

计算下列定积分

f[1,4]lnx/√xdx f[0,1]xarctanxdx

Answer 1

∫(1到4) lnx/√x
=∫(1到4)lnx d(2√x)
=lnx*2√x-∫(1到4) 2√x d(lnx)
=2√xlnx-4√x (1到4)
=(2√4*ln4-4√4)-(2√1*ln1-4√1)
=8ln2-8-0+4
=8ln2-4

∫xarctanxdx=1/2 ∫arctanxdx^2
=1/2[x^2arctanx|(0,1)-∫(0,1)x^2/(1+x^2)dx]
=1/2[π/4-∫(0,1)1-1/(1+x^2)dx]
=1/2[π/4-∫(0,1)dx+∫(0,1)1/(1+x^2)dx]
=1/2[π/4-x|(0,1)+arctanx|(0,1)]
=π/4-1/2

希望能帮到你, 望采纳. 祝学习进步

Answer 2

把分子写成3x^4-2x^2+1-x^2+x^2=(3x^4-3x^2)+(x^2+1)=3x^2(x^2-1)+(x^2-1)+2=(3x^2+1)(x^2-1)+2
再分开就行了