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

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt x\left(\sqrt[3]x+1\right)^2}. $$

Explanation of Method

Since the integrand mixes a square root and a cube root, substituting $u=x^{1/6}$ (the least common denominator of $2$ and $3$) rationalizes both radicals simultaneously.

Substitute $u=x^{1/6}$.
Simplify to $6\int u^2/(u^2+1)^2\,du$.
Integrate using the standard split $u^2=(u^2+1)-1$.
Back-substitute.

Step 1 · Substitute

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

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

Step 2 · Split and integrate

$$ \frac{u^2}{(u^2+1)^2}=\frac1{u^2+1}-\frac1{(u^2+1)^2} $$

$$ 6\int\left(\frac1{u^2+1}-\frac1{(u^2+1)^2}\right)du = 6\arctan u - 6\left[\frac u{2(u^2+1)}+\frac12\arctan u\right] = 3\arctan u - \frac{3u}{u^2+1}. $$

Step 3 · Back-substitute

$$ I = 3\arctan\left(\sqrt[6]x\right) - \frac{3\sqrt[6]x}{\sqrt[3]x+1} + C. $$

Summary Table

Piece	Antiderivative
$1/(u^2+1)$	$\arctan u$
$-1/(u^2+1)^2$	$-\dfrac u{2(u^2+1)}-\dfrac12\arctan u$

Additional Notes

The domain requires $x>0$.
The result was verified numerically at $x=1$ ($u=1$): the chain-rule derivative evaluates to $\tfrac14$, matching the integrand $1/[1\cdot(1+1)^2]=\tfrac14$ there.

Result

The solution to the integral is:

$$ I = 3\arctan\left(x^{1/6}\right) - \frac{3x^{1/6}}{x^{1/3}+1} + C. $$