Integrals Involving Roots — Question 1

Question 1 – Solution

Step 1: Substitute $u=\sqrt{x}$. Then $$x=u^2,\qquad dx=2u\,du.$$ Step 2: Rewrite the integral. $$\begin{align*} \int\frac{dx}{1+\sqrt{x}} &=\int\frac{2u}{1+u}\,du. \end{align*}$$ Step 3: Divide the numerator. $$\frac{u}{u+1}=\frac{(u+1)-1}{u+1}=1-\frac1{u+1}.$$ Step 4: Integrate and return to $x$. $$\begin{align*} 2\int\left(1-\frac1{u+1}\right)du &=2u-2\ln|u+1|+C\\ &=2\sqrt{x}-2\ln(1+\sqrt{x})+C. \end{align*}$$ $$\boxed{2\sqrt{x}-2\ln(1+\sqrt{x})+C}$$