1 ∫(上限2,下限0)1+x分之1*dx
=ln(1+x)|(上限2,下限0)
=ln(1+2)-ln(1+0)
=ln3-ln1
=ln3
2 ∫(上限1,下限0)x²+1分之3x的4次方+3x²+5*dx
=∫(上限1,下限0)[3x²+5/(x²+1)]*dx
=[x的立方+5arctanx]|(上限1,下限0)
=(1+5派/4)-(0+0)
=1+(5派/4)
1 ∫(上限2,下限0)1+x分之1*dx
=ln(1+x)|(上限2,下限0)
=ln(1+2)-ln(1+0)
=ln3-ln1
=ln3
2 ∫(上限1,下限0)x²+1分之3x的4次方+3x²+5*dx
=∫(上限1,下限0)[3x²+5/(x²+1)]*dx
=[x的立方+5arctanx]|(上限1,下限0)
=(1+5派/4)-(0+0)
=1+(5派/4)