Given: #f(x) = x + 2/x - 3/x^2#
Required: Antiderivative of #f(x)#
Theorems, Definitions and Principles:
A function #F(x)# is an antiderivative or an indefinite integral of the function #f(x)# if the derivative #F' = f#. We use the notation
#color(brown)(F(x)=intf(x)dx)#
to indicate that F is an indefinite integral of f. Using this notation, we have #color(brown)(F(x)=intf(x)dx)# if and only if #color(brown)(F'(x)=f(x))# or #color(brown)(f(x) = d/(dx)[intf(x)dx])#
Solution Strategy: Apply the above theorem
#f(x) = x + 2/x - 3/x^2#
thus the antiderivative #F(x)= intf(x)=int(x + 2/x - 3/x^2) dx#
Now to integrate apply linearity and power rule and knowledge of standard integrals:
power rule #=> intx^ndx = 1/(n+1)x^(n+1)#
#F(x)= intx dx + int2/x dx - int3/x^2 dx # Apply linearity
#F_1(x) = intx dx= x^2/2+C_1 " :"# Apply power rule #n=2#
#F_2(x) = 2int1/x dx= 2lnx +C_2" :"# linearity and standard integral
#F_1(x)=3int1/x^2 dx= -3/x+C_3" :"#Apply power rule #n=-2#
Regroup all together and the antiderivative is:
#F(x) =x^2/2 + 2lnx + 3/x+C#
#C# consolidates the other constants #C_1, C_2, C_3#
Check: #(dF(x))/(dx)= cancel2x/cancel2+2/x-3/(x^2)=x+2/x+3/x^2#
