My previous answer:
integral((3x^2+2x)dx)=integral(3x^2dx)+integral(2xdx) for 1<x<3: [x^3+x^2](1<x<3)=(3^3+3^2)-(1^3+1^2)=36-2=34.
Now your comments. First of all the 3 and 2. When you integrate x^n with respect to x, the formula you use is x^(n+1)/(n+1) where n is the power of x, and n cannot be -1. So take 3x^2. Constants are unaffected by integration, so 3 is not affected. That means integral of 3x^2 is the same as 3 times integral of x^2, and integral(2xdx)=2integral(xdx). Using the formula and n=2, x^2 integrated is x^3/3. The 3 in 3x^2 is unaffected so we carry it with us as a multiplier when we integrate and we get 3*x^3/3. The 3 cancels with the 3 in the denominator of x^3/3 so we're left with just x^3. The same happens to the 2 when we integrate 2x, this time n=1. Integrate x and we get x^2/2, put the unchanged 2 in front as a multiplier and we have 2*x^2/2=x^2, because the twos cancel as the threes did. The other thing about integrals is that you can split the integral into two because the + sign (or a minus sign) is a separator as well as an operator. That's why integral((3x^2+2x)dx) is the same as integral(3x^2dx)+integral(2xdx)=3integral(x^2dx)+2integral(xdx).
You can't do this with multiply or divide, so if A and B are integrals (containing a variable, rather than just constants), we can't do integral(A*B)=integral(A)*integral(B). The same applies to divide. It's easy to prove this using the two integrals 3x^2 and 2x. Multiply them together: 6x^3 which when integrated is 6x^4/4=3x^4/2 and this is not the same as x^3 times x^2=x^5.
The final bit is substitution. You subtract the lower limit result from the upper limit result, substituting x=1 and x=3 into the results of the integration: x^3+x^2=27+9=36 when x=3 and 1+1=2 when x=1.
I hope that this explanation helps you to understand what's going on.