1, first-order linear differential equation, direct set of formulas

y=e^-∫ 1dx=π/2

3. let x=tanu, then dx = (secu) 2du, x: 0-> 1, u: 0-> π/4

Original formula = ∫1/secu * (secu) 2du = ∫ secudu = ln | secu+tanu | [0-> π/4]

=ln(√2+ 1)

4, partial integral:

Original formula = xarcsinx-∫ x/√ (1-x 2) dx = xarcsinx-1/2 ∫1√ (1-x 2) d (x

= xarcsinx+√ (1-x 2) is substituted into the upper and lower limits for subtraction.

= 1/2*π/6+√3/2- 1

=π/ 12++√3/2- 1

5、y'=3x^2cos3x-3x^3sin3x

Then dy = (3x2cos3x-3x3sin3x) dx.

lim[x-& gt； ∞] ( 1-2/x)^(-x)

= lim[x->； ∞] [( 1-2/x)^(-x/2)]^2

= e 2 seconds important limit

lim[x-& gt； ∞] xsin( 1/x)

= lim[x->； ∞] sin( 1/x)/( 1/x)

= 1

8. y =1/x 2, then y' =-2/x 3, and substitute x= 1/π to get y' | (x =1/π) =-2π 3.

9. The result of definite integral is constant, and the derivative of constant is 0, so the result of this problem is 0.