Question

怎样求定积分？

求解？怎样求定积分？

Answer 1

Answer 2

解：1.设√(1+x)=t，则x=t²-1，dx=2tdt
故∫<0,3>xdx/[1+√(1+x)]=∫<1,2>(t²-1)*2tdt/(1+t)
=2∫<1,2>(t²-t)dt
=2(t³/3-t²/2)│<1,2>
=2(8/3-2-1/3+1/2)
=5/3；
2.设x=3tant，则dx=3sec²tdt
故∫<0,4>√(x²+9)dx
=∫<0,arctan(4/3)>3sect*3sec²tdt
=9∫<0,arctan(4/3)>costdt/(cos²t)²
=9∫<0,arctan(4/3)>d(sint)/(1-sin²t)²
=(9/4)∫<0,arctan(4/3)>[1/(1+sint)+1/(1-sint)+1/(1+sint)²+1/(1-sint)²]d(sint)
=(9/4)[ln(1+sint)-ln(1-sint)-1/(1+sint)+1/(1-sint)]│<0,arctan(4/3)>
=(9/2)[ln((1+sint)/cost)+sint/cos²t]│<0,arctan(4/3)>
=(9/2)[ln((1+4/5)/(3/5))+(4/5)/(3/5)²]
=9ln3/2+30

Answer 3

第一个根号代换令t等以根号x+1；
第二 x=3tant