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

1 Answer

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.