Illustration of how Mathematica works

Solving ODEs

DSolve[y'[x]+2/x*y[x]==1/x^4*y[x]^4,y[x],x]

Sometimes Mathematica writes things in an ugly way - 
note that it wrote (((-3))^(1/3)) and (((-1))^(2/3)) which is not real numbers.  
Here C[1] is the arbitrary constant; if the equation were of order 2, 
then we would have two arbitrary constants, C[1] and C[2].

DSolve[y'[x]==2*(y[x]+5)*Exp[x],y[x],x]

DSolve[y'[x]==(Sqrt[y[x]]-y[x])/Sin[x],y[x],x]

This is certainly not a pretty way to write the result.

Checking that a function is a solution of the ODE

y1[x_]=(3*x^3/(1+C*x^9))^(1/3)

Simplify[y1'[x]+2/x*y1[x]-1/x^4*y1[x]^4]

y2[x_]=C*Exp[2*Exp[x]]-5

Simplify[y2'[x]-2*(y2[x]+5)*Exp[x]]

y3[x_]=(1+C/Sqrt[Tan[x/2]])^2

Simplify[y3'[x]-(Sqrt[y3[x]]-y3[x])/Sin[x]]

Mathematica needs some help, so I will simplify these expressions one by one:

Simplify[y3'[x]]

Simplify[-(Sqrt[y3[x]]-y3[x])/Sin[x]]

I will help Mathematica by telling it that Sqrt[A^2]=A - which, of course, 
is not always true, so it is good that Mathematica is cautious!

Simplify[Csc[x] ((((1+(C/(Sqrt[Tan[(x/2)]]))))^2)-(1+(C/(Sqrt[Tan[(x/2)]]))))]

Comparing the expressions above, we see that y3[x] is indeed a solution of the ODE

Solving integrals

Integrate[1/(v^4-3*v),v]

Integrate[1/(Sqrt[y]-y),y]

Integrate[1/Sin[x],x]

Integrate[Sqrt[Tan[x/2]]/2/Sin[x],x]