Evaluate the integral: I = integral dx/(sqrt x+ sqrt x)

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt x+\sqrt[3]x}. $$

Step 1 · Substitute

Let $u=x^{1/6}$, so $x=u^6$, $dx=6u^5\,du$, $\sqrt x=u^3$, $\sqrt[3]x=u^2$:

$$ I = \int\frac{6u^5\,du}{u^3+u^2} = 6\int\frac{u^3}{u+1}\,du. $$

Step 2 · Divide

$$ \frac{u^3}{u+1} = u^2-u+1-\frac1{u+1}. $$

Step 3 · Integrate and back-substitute

$$ 6\int\left(u^2-u+1-\frac1{u+1}\right)du = 2u^3-3u^2+6u-6\ln|u+1|+C. $$

$$ I = 2\sqrt x - 3\sqrt[3]x + 6\sqrt[6]x - 6\ln\left(\sqrt[6]x+1\right) + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq0$.
• The result was verified numerically at $x=1$ ($u=1$): the chain-rule derivative evaluates to $0.5$, matching the integrand $1/(1+1)=0.5$ there.