Question

What is `int1/(2+sqrtx)dx` using substitution u=sqrtx ?

Answer 1

`int \ 1/(2+sqrt(x)) \ dx = 2 sqrt(x)-4ln|2+sqrt(x)| + C`

Explanation:

Let:

`I = int \ 1/(2+sqrt(x)) \ dx`

If we perform the suggested change of variable then we have;

`u =sqrt(x) => (du)/dx = 1/(2sqrt(x)) = 1/(2u)`

Then performing the substitution we get:

`I = int \ 1/(2+u) \ 2u \ du`
`\ \ = 2 \ int \ u/(2+u) \ du`
`\ \ = 2 \ int \ (2+u-2)/(2+u) \ du`
`\ \ = 2 \ int \ 1-2/(2+u) \ du`

We can now integrate to get:

`I = 2 {u-2ln|2+u|} + C`
`\ \ = 2 u-4ln|2+u| + C`

Restoring the substitution we get:

`I = 2 sqrt(x)-4ln|2+sqrt(x)| + C`

Having shown this solution, however, I would have used the substitution:

`u =2+sqrt(x) => (du)/dx = 1/(2sqrt(x)) = 1/(2(u-2))`

Then performing the substitution we get:

`I = int \ 1/u \ 2(u-2) \ du`
`\ \ = 2 \ int \ 1-2/u \ du`

We can now integrate to get:

`I = 2{ u - 2ln |u| } + C`
`\ \ = 2u - 4ln |u| + C`
`\ \ = 2(2+sqrt(x) ) - 4ln |2+sqrt(x) | + C`
`\ \ = 4+2sqrt(x) - 4ln |2+sqrt(x) | + C`
`\ \ = 2sqrt(x) - 4ln |2+sqrt(x) | + C_1`, as above