Question

What is `int_(2)^(3) (x-1)/(x^3)+x^2dx`?

Answer 1

`int_2^3 ((x−1)/x^3+x^2)dx=463/72 approx 6.4306`

Explanation:

First of all, let's split this integral (we can do this thanks to a property called linearity):
`int_2^3 ((x−1)/x^3+x^2)dx=int_2^3 x/x^3 dx - int_2^3 1/x^3 dx+int_2^3 x^2 dx=int_2^3 1/x^2 dx - int_2^3 1/x^3 dx+int_2^3 x^2 dx`

Now we can find a primitive function for each of the three integrand functions and evaluate them in the integration extrema:
`int_2^3 1/x^2 dx=int_2^3 x^(-2) dx=[x^{-1}/-1]_2^3=[-1/x]_2^3 =-1/3+1/2=1/6`
`int_2^3 1/x^3 dx=int_2^3 x^(-3) dx=[x^{-2}/-2]_2^3=[-1/(2x^2)]_2^3 =-1/18+1/8=5/72`
`int_2^3 x^2 dx=[x^3/3]_2^3=9-8/3=19/3`

Now we just sum up the three results (the second one has to be subtracted!) and get
`int_2^3 ((x−1)/x^3+x^2)dx=1/6-5/72+19/3=463/72`

Note: To calculate the three primitives, we used the same rule: all the functions of type `x^alpha` where `alpha in RR` with `alpha ne -1` admit `x^{alpha+1}/(alpha+1)` as a primitive function. If `alpha = -1` the primitive function is the natural logarithm.