Question

Integrate the following with respect to x :

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

Answer 1

`int sqrt(x)/(1 + sqrt(x))  "d"x`

Put `1 + sqrt(x)` = u

`1 + x^(1/2)` = u

`(0 + 1/2 x^(1/2 - 1))  "d"x` = du

`1/2  x^(- 1/2)  "d"x` = du

`1/(2 x^(1/2))  "d"x` = du

`1/sqrt(x)  "d"x` = 2 du

`int sqrt(x)/(1 + sqrt(x))  "d"x = int (sqrt(x) * sqrt(x))/(sqrt(x) * (1 + sqrt(x))  "d"x`

`int sqrt(x)/(1 + sqrt(x))  "d"x = int x/((1 + sqrt(x))) * 1/sqrt(x)  "d"x`  .......(1)

`1 + sqrt(x)` = u

`sqrt(x)= "u" - 1`

x = `("u" - 1)^2`

(1) ⇒ `int sqrt(x)/(1 + sqrt(x))  "d"x = int ("u" - 1)^2/"u"  2"du"`

= `2 int ("u"^2 - 2"u" + 1)/"u"  "du"`

= `2 int ("u"^2/"u" - (2"u")/"u" + 1/"u")  "du"`

= `2 int ("u" - 2 + 1/"u")  "du"`

= `2[int "u"  "du" - int 2  "du" + int 1/"u"  "du"]`

= `2 ["u"^2/"u" - 2"u" + log|"u"|] + "c"`

= `2[(1 + sqrt(x))^2/2 - 2(1 + sqrt(x)) + log|1 + sqrt(x)|] + "c"`

`int sqrt(x)/(1 + sqrt(x))  "d"x = (1 + sqrt(x))^2 - 4(1 + sqrt(x)) + 2log|1 + sqrt(x)| + "c"`