Question

Integrate thefollowing question :
integrate 1/1+rootx

Answer 1

Answer:

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

Lets use substitution, x = t².

So, dx = 2tdt

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

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

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

Add and Subtract 1 on the numerator.

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

$\sf \to 2 \left( \displaystyle \int \sf \dfrac{t+1}{1+t}dt - \int \sf \dfrac{1}{1+t}dt \right)$

$\sf \to 2 \left( \displaystyle \int \sf dt - \int \sf \dfrac{1}{1+t}dt \right)$

Formulas I am about to use:
$\to \boxed{\bf{\displaystyle \int \bf dx = x}}$

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

Our Integral becomes:

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

x = t², so t = √x

$\sf \to 2 \left( \sqrt{x} - ln(\sqrt{x} +1 \right))$

Answer:
$\boxed{\pink{\bf \to 2 \left( \sqrt{x} - ln(\sqrt{x} +1 \right))}}$