Question

Question #07873

Answer 1

`int(1)/(x+sqrtx)dx = 2 ln(sqrt(x) + 1) + C`, with `x>= 0` and where `C` is the constant of integration.

Explanation:

We want to compute the antiderivative of `int(1)/(x+sqrtx)dx`.
Since we have `sqrt(x)`, we'll assume `x>=0`.

We will use this change of variable :
`x = phi(t) = t^2`

`int(1)/(x+sqrtx)dx` = `int(1)/(t^2+sqrt(t^2))*dot ((t^2))dt`

`= int(2t)/(t^2+t)dt = 2 int 1/(t+1) dt = 2 ln(|t+1|) + C.`

Since `x = t^2`,`t = sqrt(x)` for `x >= 0` (which was supposed above).

So `int(1)/(x+sqrtx)dx = 2 ln(|sqrt(x) + 1|) + C = 2 ln(sqrt(x) + 1) + C`, where `C` is the constant of integration.