Question

How do you integrate `int 1/(sqrtx(1+sqrtx)^2dx` from [1,9]?

Answer 1

`int_1^9 1/(sqrtx(1+sqrtx)^2)dx=1/2`

Explanation:

`I=int_1^9 1/(sqrtx(1+sqrtx)^2)dx`

We will use the substitution `u=1+sqrtx`. Differentiating this reveals that `du=1/(2sqrtx)dx`. We can rewrite the integral to make both of these evident:

`I=2int_1^9(1+sqrtx)^(-2)(1/(2sqrtx)dx)`

Before switching from `dx` to `du`, change the variables of `x=1` and `x=9` by plugging them into `u=1+sqrtx`.

`x=1` becomes `u=1+sqrt1=2` and `x=9` becomes `u=1+sqrt9=4`. Then:

`I=2int_2^4u^-2du`

Using `intu^ndu=u^(n+1)/(n+1)`:

`I=2[u^(-1)/(-1)]_2^4=2[-1/u]_2^4=2(-1/4-(-1/2))`

`I=2(-1/4+1/2)=2(1/4)=1/2`