Integrals Involving Roots — Question 8

Question 8

Evaluate $$\int_0^1\frac{dx}{\sqrt{x}+\sqrt[3]{x}}$$ using a substitution that clears both radicals.

Question 8 – Solution

Step 1: Choose a power that clears both radicals. Since $\operatorname{lcm}(2,3)=6$, let $x=u^6$. Then $$\sqrt{x}=u^3,\qquad \sqrt[3]{x}=u^2, \qquad dx=6u^5\,du.$$ The bounds remain $0$ and $1$ because $u=x^{1/6}$. Step 2: Rewrite and simplify. $$\begin{align*} I&=6\int_0^1\frac{u^5}{u^3+u^2}\,du\\ &=6\int_0^1\frac{u^5}{u^2(u+1)}\,du =6\int_0^1\frac{u^3}{u+1}\,du. \end{align*}$$ Step 3: Divide the polynomials. $$\frac{u^3}{u+1}=u^2-u+1-\frac1{u+1}.$$ Step 4: Integrate and evaluate the bounds. $$\begin{align*} I&=6\int_0^1\left(u^2-u+1-\frac1{u+1}\right)du\\ &=6\left[\frac{u^3}{3}-\frac{u^2}{2}+u-\ln(u+1)\right]_0^1\\ &=6\left(\frac13-\frac12+1-\ln2\right)\\ &=6\left(\frac56-\ln2\right)=5-6\ln2. \end{align*}$$ $$\boxed{I=5-6\ln2}$$