# What is the integral: `int ` `1/(1+sqrt x) dx` ?

*print*Print*list*Cite

### 2 Answers

The integral `int 1/(1+sqrt x) dx` has to be derived.

`int 1/(1+sqrt x) dx`

Let `x = u^2`

`(dx)/(du) = 2u`

`dx = 2u*du `

`int 1/(1+sqrt x) dx`

= `int 1/(1+u) * 2u du`

Let `1 + u = y => dy = du`

=> `int 2*(y - 1)/y dy`

=> `2*int 1 - 1/y dy`

=> `2*(y - ln y)`

Substitute `y = 1 + u`

=> `2*(1 + u - ln(1 + u))`

Substitute `u = sqrt x`

=> `2*(1 + sqrt x - ln(1 + sqrt x)) + C`

**The required integral `int 1/(1+sqrt x) dx = 2*(1 + sqrt x - ln(1 + sqrt x)) + C` **

`int 1/(1+sqrt(x))dx` `x=t^2` `dx=2tdt`

`2int t/(1+t) dt=2int(1+t-1)/(1+t) dt =` `2int dt -2int1/(1+t) dt=2t -log(1+t)^2 +c`

`int 1/(1+sqrt(x)) dx=2sqrt(x)-log|1+2sqrt(x)+x|+c`