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The substitution $x = 1 + e^u$ simplifies the integrand because its derivative $dx = e^u du$ exactly matches the $e^u du$ term in the original integral. This allows the complex expression $e^u \sqrt{1+e^u} du$ to be rewritten as the simpler radical $\sqrt{x} dx$.

Conditions: The integrand contains $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The differential is $dx = e^u du$.

The substitution $x = 1 + e^u$ simplifies the integrand because its derivative $dx = e^u du$ exactly matches the $e^u du$ term in the original integral. This allows the complex expression $e^u \sqrt{1+e^u} du$ to be rewritten as the simpler radical $\sqrt{x} dx$.

Conditions: The integrand contains $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The differential is $dx = e^u du$.

To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

Conditions: The curve is defined by the function $f(x) = x^{3/2}$.; The interval of integration is $[0, 32/9]$.; The arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ is applicable.

To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

Conditions: The curve is defined by the function $f(x) = x^{3/2}$.; The interval of integration is $[0, 32/9]$.; The arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ is applicable.