f(x) = 1/(1+ sqrtx)

Let us use the substitution method to solve:

let x= u^2

==> dx = 2u du

Now we will substitute:

intg f(x) = intg (1/(1+u) 2udu

= intg (2u/(1+u) du

Now simplify using polynomial division:

=> intg f(x) = intg ( 2 - 2/(u+1) du

= intg 2 du - 2 intg (1/u+1) du

= 2u - 2 ln(u+1) + C

Now subsitute with u= sqrtx

==>** intg f(x) = 2sqrtx - 2ln(sqrtx + 1) + C**

We'll have to integrate the indefinite integral :

Int dx/(1 + sqrtx)

We'll substitute 1 + sqrt x = t

sqrt x = t-1

We'll differentiate both sides and we'll get:

dx/2sqrt x = dt

dx = 2sqrt x*dt

dx = 2(t-1)dt

We'll re-write the integral in the variable t:

Int 2(t-1)dt/t

We'll apply the property of integral to be additive:

Int 2(t-1)dt/t = 2 int tdt/t - 2 Int dt/t

Int 2(t-1)dt/t = 2 Int dt - 2 ln |t|

Int 2(t-1)dt/t = 2t - 2 ln |t|

**Int dx/(1 + sqrtx) = 2*(1 + sqrt x) - ln (1 + sqrt x)^2 + C**

To integrate f(x) = 1/sqrtx.

We put sqrtx = t.

d/dx(sqrtx) = dt.

dx/2sqrtx = dt.

dx = 2t dt.

Therefore Int {1/(1+sqrtx)} dx =Int 2tdt/1+t

Int {1/(1+sqrtx)} dx= Int {2 -2/(1+t)}dt

Int {1/(1+sqrtx)} dx = 2t - 2log(1+t)

Int {1/(1+sqrtx)} dx = 2sqrtx - 2log(1+sqrtx) +Const.

Therefore Int {1/sqrtx} dx = 2sqrtx - 2log(1+sqrtx) + Const.