∫(x？ +seconds? x)dx=( 1/3)x？ +tanx+C

∫[ 1/(2x-3)]dx=( 1/2)∫[ 1/(2x-3)]d(2x-3)=( 1/2)ln∣2x-3∣+c

∫sin(x/2)dx = 2∫sin(x/2)d(x/2)=-2cos(x/2)+C

∫dx/( 1+√x)； Let x=u, then x = u? ，dx = 2udu

So the original formula = 2 ∫ udu/(1+u) = 2 ∫ [1-1(1+u)] du = 2 [u-ln1.

∫xe^(3x)dx=( 1/3)∫xd[e^(3x)]=( 1/3)[xe^(3x)-∫e^(3x)dx]=( 1/3)e^(3x)-∫e^(3x)d(3x)

=( 1/3)e^(3x)-e^(3x)+c=-(2/3)e^(3x)+c

-2，- 1∫( 1+2x)？ dx=-2，- 1( 1/2)∫( 1+2x)？ d( 1+2x)=( 1/6)( 1+2x)？ -2,- 1=-( 1/6)+9/2= 13/3

0， 1/2∫arcsinxdx =[xarcsinx+√( 1-x？ )]0, 1/2=π/6+(√3/2)- 1

0， 1/4∫xdx/√( 1-2x)=0， 1/4-∫xd √( 1-2x)=-[x √( 1-2x)-∫√( 1-2x)dx]0， 1/4

=-[x √( 1-2x)+( 1/2)∫√( 1-2x)d( 1-2x)]0， 1/4

=-[x √( 1-2x)+( 1/3)x √( 1-2x)+√( 1-2x)？ ]0, 1/4

=-[ 1/(4√2)+ 1/( 12√2)+ 1/(2√2)]+ 1=( 12+5√2)/ 12

(1). Find the differential equation 2xy+x? General solution of y' = (1-x) y.

Solution: x? Y'=y-3xy, that is, (3xy-y)dx+x? dy=0...........( 1);

Where P=3xy-y and Q=x? ; ? P/？ y = 3x- 1≦？ Q/？ X=2x, so the original equation is not a fully differential equation.

But (1/Q) (? P/？ You-? Q/？ x)=( 1/x？ )(3x- 1-2x)=(x- 1)/x？ =G(x) is a function of x,

So there is an integral factor μ (x) = e ∫ g (x) dx = e ∫ [(1/x)-(1/x? )]dx=e^(lnx+ 1/x)=xe^( 1/x)；

Multiply μ(x) by [(3xy-y) Xe (1/x)] dx+[x? e^( 1/x)]dy=0..........(2)

At this time, p = (3xy-y) xe (1/x); Q=x？ e^( 1/x)；

P/？ y=(3x- 1)xe^( 1/x)=？ Q/？ x=3x？ e^( 1/x)-xe^( 1/x)=(3x- 1)xe^( 1/x)； Therefore, (2) is a fully differential equation.

(2) On the left is the function u(x, y) = ∫ [(3xy-y) xe (1/x)] dx = yx? Total differential of e (1/x). Because:

Du = (? u/？ x)dx+(？ u/？ y)dy=y[(3x？ -x]e^( 1/x)dx+x？ E (1/x)] dy = (2) left.

What about implicit functions? yx？ E (1/x) = c is the general solution of the original equation.

(2) Find the general solution of the differential equation y''+2y'+y = e (-x).

Solution: The characteristic equation of homogeneous equation y''+2y'+y=0 is R? +2r+ 1=(r+ 1)？ =0, with multiple roots r? =r？ =- 1;

So the general solution of homogeneous equation is y=(C? +C？ x)e^(-x)

Let's find a special solution y *；; Using the undetermined coefficient method: let the special solution be y*=ax? e^(-x)

(y*)'=2axe^(-x)-ax？ e^(-x)=(2ax-ax？ )e^(-x)

(y*)''=(2a-2ax)e^(-x)-(2ax-ax？ )e^(-x)=(2a-4ax+ax？ )e^(-x)

Substitute into the original formula (2a-4ax+ax? )e^(-x)+2(2ax-ax？ )e^(-x)+ax？ e^(-x)=e^(-x)

(2a-4ax+ax？ )+4ax-2ax？ +ax？ = 1

That is, 2a= 1, so a= 1/2, that is, the special solution is y*=( 1/2)x? e^(-x)

So the general solution of the original equation is: y=(C? +C？ x)e^(-x)+( 1/2)x？ e^(-x)=[C？ +C？ x+( 1/2)x？ ]e^(-x)