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The antiderivative of $\sqrt{u}$ is found by rewriting the square root as a fractional power, $u^{1/2}$, and then applying the power rule for integration. The power rule states that $\int u^n du = \frac{u^{n+1}}{n+1}$.

Conditions: The integrand is $\sqrt{u}$, which is equivalent to $u^{1/2}$.; The power rule for integration is applicable.; $u \ge 0$ for the real-valued square root form.

The antiderivative of $\sqrt{u}$ is found by rewriting the square root as a fractional power, $u^{1/2}$, and then applying the power rule for integration. The power rule states that $\int u^n du = \frac{u^{n+1}}{n+1}$.

Conditions: The integrand is $\sqrt{u}$, which is equivalent to $u^{1/2}$.; The power rule for integration is applicable.; $u \ge 0$ for the real-valued square root form.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.

The substitution $u = 1 + \frac{9}{4}x$ is chosen because it is exactly the expression under the radical in the arc-length integral $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$. By setting $u$ to this inner expression, the complicated integrand simplifies to $\sqrt{u}$, which is much easier to integrate using the power rule.

Conditions: The integral to evaluate is $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$.; The integrand contains a composite expression under a square root.

The substitution $u = 1 + \frac{9}{4}x$ is chosen because it is exactly the expression under the radical in the arc-length integral $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$. By setting $u$ to this inner expression, the complicated integrand simplifies to $\sqrt{u}$, which is much easier to integrate using the power rule.

Conditions: The integral to evaluate is $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$.; The integrand contains a composite expression under a square root.