Question

How do you evaluate the definite integral `int (3/x^2-1)dx` from [1,2]?

Answer 1

I found: `1/2`

Explanation:

Let us write our integral in a slight different way to make the integraton easier:
`int3x^-2dx-int1dx=`
now we integrate using our usual rules of integration to get:
`=3x^(-2+1)/(-2+1)-x=-3x^-1-x=-3/x-x`
we didn't add the constant because we now apply the limits of integration `[1,2]` evaluating our result for `x` in both points and subtracting the values:
when `x=2` we get:
`-3/2-2=-7/2`

when `x=1` we get:
`-3/1-1=-4`

Subtract them: `(-7/2)-(-4)=(-7+8)/2=1/2`

so we get that:
`int_1^2(3/x^2-x)dx=1/2`