=liml[ln(e^(2x)+ 1)/x

= lim2 (e (2x)/(e (2x)+ 1) (Robida's law)

=2

2、∵(sinx)^5dx=-(sinx)^4d(cosx)=-[( 1-(cosx)^2]^2d(cosx)，

∴ Original formula =-15 [cosx-(2/3) (cosx) 3+(1/5) (cosx) 5] 丨 (x=0, π/2).

=8

Note: this question can also be directly obtained by formula.

3. The integration area D={(x, y) è-1≤x≤ 1,-1≤y≤x},

∴k3=-3∫(- 1, 1)dx∫(- 1,x)[y+yxe^(x^2+y^2)/2]dy。

And ∫ (- 1, x) [y+yxe (x 2+y 2)/2] dy.

= [( 1/2) y 2+xe (x 2+y 2)/2] è (y=- 1, x)

=( 1/2)(x^2- 1)+xe^(x^2)-xe^(x^2+ 1)/2；

In the integral interval x∈[- 1], xe (x 2)-xe (x 2+1)/2 is odd function, and its integral is 0.

∴ Original formula = (-3/2) ∫ (- 1, 1) (x 2- 1) dx = 2.

4, let y'-y=0, then dy/y=dx, y * = ce x.

Let y = v (x) e x and bring it into the original equation. V' (x) = (1-x 2) e (-x). The integral is v (x) = (x 2+2x+ 1) e (-x)+c,

∴y=(x+ 1)^2+ce^x。

In addition, f(x)=y is a quadratic function, ∴c=0.

Note: This problem can also be directly solved by the general solution formula of the first-order linear equation.

∴ Original formula =f( 1)=4.

5. Transform D={(x, y) 乸 y≤x≤π/6, 0≤y≤π/6} into D={(x, y) 乸 0≤y≤x, 0≤x≤π/6},

Exchange integration orders,

The original formula =2∫(0, π/6)(cosx/x)dx∫(0, x)dy.

=2∫(0, π/6)cosxdx=2sinx 丨 (x=0, π/6)

= 1。