Evaluate the integral: I = integral (x²sqrt x)/(x+ sqrt

Evaluate the integral: $$ I = \int \frac{x^2\sqrt x}{x+\sqrt x}\,dx. $$

Explanation of Method

Substituting $u=\sqrt x$ turns the integrand into $u^5/(u+1)$, an improper rational function resolved by polynomial long division.

Let $u=\sqrt x$, so $x=u^2$, $dx=2u\,du$, and $x+\sqrt x=u^2+u=u(u+1)$:

$$ I = \int\frac{u^4\cdot u}{u(u+1)}\cdot2u\,du = 2\int\frac{u^5}{u+1}\,du. $$

$$ \frac{u^5}{u+1} = u^4-u^3+u^2-u+1-\frac1{u+1}. $$

$$ 2\int\left(u^4-u^3+u^2-u+1-\frac1{u+1}\right)du = \frac{2u^5}5-\frac{u^4}2+\frac{2u^3}3-u^2+2u-2\ln|u+1|+C. $$

$$ I = \frac{2x^{5/2}}5 - \frac{x^2}2 + \frac{2x^{3/2}}3 - x + 2\sqrt x - 2\ln\left(\sqrt x+1\right) + C. $$

Piece	Antiderivative
$u^4-u^3+u^2-u+1$	$\tfrac{u^5}5-\tfrac{u^4}4+\tfrac{u^3}3-\tfrac{u^2}2+u$
$-1/(u+1)$	$-\ln\|u+1\|$

The domain requires $x\geq0$.
The result was verified numerically at $x=1$ ($u=1$): the derivative of the boxed expression evaluates to $\tfrac12$, matching the integrand $1^2\cdot1/(1+1)=\tfrac12$ there.

The solution to the integral is:

$$ I = \frac{2x^2\sqrt x}5 - \frac{x^2}2 + \frac{2x\sqrt x}3 - x + 2\sqrt x - 2\ln\left(\sqrt x+1\right) + C. $$