Integrals Involving Roots — Question 9

Question 9 – Solution

Step 1: Differentiate with the chain rule. $$\begin{align*} F'(x)&=\frac23\cdot\frac32(1+x^2)^{1/2}\cdot(2x)\\ &=2x\sqrt{1+x^2}. \end{align*}$$ Step 2: Reverse the process by integration. $$\int 2x\sqrt{1+x^2}\,dx.$$ Let $$u=1+x^2,\qquad du=2x\,dx.$$ Then $$\begin{align*} \int 2x\sqrt{1+x^2}\,dx &=\int u^{1/2}\,du\\ &=\frac{u^{3/2}}{3/2}+C\\ &=\frac23(1+x^2)^{3/2}+C. \end{align*}$$ $$\boxed{\int2x\sqrt{1+x^2}dx=F(x)+C}$$