Integrals Involving Roots — Question 5

Question 5 – Solution

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