Question

Integrate the following
Integrate 1/1+rootx

Answer 1

Answer:

This is the answer to the question

Answer 2

Answer:

$\sf \to \displaystyle \sf \int \dfrac{1}{1+\sqrt{x}} dx$

Substitute x=t², dx=2tdt.

$\sf \to \displaystyle \sf \int \dfrac{2t}{1+\sqrt{t^2}} dt$

$\sf \to 2 \displaystyle \sf \int \dfrac{t}{1+t} dt$

Add and subtract 1 in the numerator.

$\sf \to 2 \displaystyle \sf \int \dfrac{t+1-1}{1+t} dt$

$\sf \to 2 \displaystyle \int \sf \dfrac{t+1}{1+t} dt - \displaystyle \int \sf \dfrac{1}{1+t} dt$

$\sf \to 2 t- 2ln(t+1)+c$

Substitute t back in the equation.

$\to \boxed{\blue{\sf 2 \sqrt{x} - 2 \: ln(\sqrt{x} +1)+c}}$

And we are done! The formulas I used while simplifying:

$\to \displaystyle \sf \int \dfrac{1}{x+(constant)}dx=ln(x+(constant)$

$\to \displaystyle \sf \int dx=x$