Question

How do you find the definite integral of `int 3sqrt(1+x^(-1))dx` from `[0,3]`?

Answer 1

`3/2(4sqrt(3)-log(2/sqrt(3)-1)+log(1+2/sqrt(3)))`

Explanation:

Writing the integrand in the form
`3int_0^3sqrt((x+1)/x)dx`
`u=(x+1)/x` `du=(1/x-(x+1)/x^2)dx`
then we get
`-3int_infty^(4/3)sqrt(u)/(1-u)^2du`
`3int_(4/3)^inftysqrt(u)/(1-u)^2du`

substituting `s=sqrt(u),ds=1/(2*sqrt(u))du`
so

`6*int_(2/sqrt(3))^infty s^2/(1-s^2)^2ds`

`6int_(2sqrt(3))^infty -1/(4(s+1))+1/(4*(s-1)^2)+1/(4(s-1))+1/(4(s-1)^2)ds`

`lim_(b to infty)3/2(-2s(s^2+1)+log(1-s)-log(1+s))_(2/sqrt(3))^infty`
and this gives

`3/2*(4sqrt(3)-log(2/sqrt(3)-1)+log(1+2/sqrt(3))`