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To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

The factor $x^3$ is grouped with $\sqrt{1+x^4}$ because the derivative of the inner function $1+x^4$ is $4x^3$. This presence of $x^3$ makes the expression $x^3\sqrt{1+x^4}$ directly integrable using substitution, which is a prerequisite for the integration by parts method.

Conditions: The integrand is $x^7\sqrt{1+x^4}$.; The method involves integration by parts.; The term $dv$ must be integrable.

The factor $x^3$ is grouped with $\sqrt{1+x^4}$ because the derivative of the inner function $1+x^4$ is $4x^3$. This presence of $x^3$ makes the expression $x^3\sqrt{1+x^4}$ directly integrable using substitution, which is a prerequisite for the integration by parts method.

Conditions: The integrand is $x^7\sqrt{1+x^4}$.; The method involves integration by parts.; The term $dv$ must be integrable.

Rewriting the integrand separates it into a factor suitable for differentiation and a factor suitable for integration. Choosing $x^4$ for differentiation reduces its power, while the remaining factor $x^3\sqrt{1+x^4}$ becomes integrable via substitution because its derivative contains the necessary $x^3$ term.

Conditions: The integrand is $x^7\sqrt{1+x^4}$.; Integration by parts is the chosen method.; A substitution is available for the second factor.

Rewriting the integrand separates it into a factor suitable for differentiation and a factor suitable for integration. Choosing $x^4$ for differentiation reduces its power, while the remaining factor $x^3\sqrt{1+x^4}$ becomes integrable via substitution because its derivative contains the necessary $x^3$ term.

Conditions: The integrand is $x^7\sqrt{1+x^4}$.; Integration by parts is the chosen method.; A substitution is available for the second factor.

The final value is $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This is obtained by evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at the converted bounds $x=3$ and $x=2$ and subtracting the results.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

The final value is $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This is obtained by evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at the converted bounds $x=3$ and $x=2$ and subtracting the results.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

These techniques are combined when the integrand can be split into a product where one factor simplifies upon differentiation (for integration by parts) and the other factor is integrable only via substitution. This often occurs with polynomials multiplied by composite functions like $\sqrt{ax^n+b}$.

Conditions: The integrand is a product of two distinct types of functions.; One part reduces nicely when differentiated.; The other part requires a u-substitution to find its antiderivative.

These techniques are combined when the integrand can be split into a product where one factor simplifies upon differentiation (for integration by parts) and the other factor is integrable only via substitution. This often occurs with polynomials multiplied by composite functions like $\sqrt{ax^n+b}$.

Conditions: The integrand is a product of two distinct types of functions.; One part reduces nicely when differentiated.; The other part requires a u-substitution to find its antiderivative.

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.

For the arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ to be valid, the function $f(x)$ must be continuous on the closed interval $[a, b]$, and its derivative $f'(x)$ must also be continuous on $[a, b]$. These conditions ensure that the curve is smooth enough for the integral to accurately represent its length.

Conditions: The curve is defined by a function $y = f(x)$.; The interval of integration is $[a, b]$.; $f(x)$ is continuous on $[a, b]$.; $f'(x)$ is continuous on $[a, b]$.

For the arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ to be valid, the function $f(x)$ must be continuous on the closed interval $[a, b]$, and its derivative $f'(x)$ must also be continuous on $[a, b]$. These conditions ensure that the curve is smooth enough for the integral to accurately represent its length.

Conditions: The curve is defined by a function $y = f(x)$.; The interval of integration is $[a, b]$.; $f(x)$ is continuous on $[a, b]$.; $f'(x)$ is continuous on $[a, b]$.

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.