int_((1)/(2))^(2)(2x^(4))/(x^(5)+1)dx

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of the function (2x^4)/(x^5 + 1) from x = 1/2 to x = 2.Simplify Integral: Simplify the integral if possible. In this case, the integral does not simplify easily, so we proceed with the given integral.Perform Substitution: Perform a substitution if applicable. Let u = x^5 + 1, then du = 5x^4 dx. We need to adjust the integral to match this substitution. To do this, we multiply and divide the integral by 5 to get the x^4 term to match du.Rewrite in terms of u: Rewrite the integral in terms of u. The integral becomes (2/5) * int((1/2)^(2), 2) (5x^4)/(x^5 + 1) dx = (2/5) * int(u(1/2), u(2)) du/u.Update Limits: Update the limits of integration. When x = 1/2, u = (1/2)^5 + 1 = 1/32 + 1 = 33/32. When x = 2, u = 2^5 + 1 = 32 + 1 = 33.Evaluate Integral: Evaluate the integral with the new limits. The integral is now (2/5) * int(33/32, 33) du/u. The integral of 1/u with respect to u is ln|u|, so we have (2/5) * [ln|u|] from u = 33/32 to u = 33.Calculate Definite Integral: Calculate the definite integral. Plug in the limits of integration: (2/5) * (ln|33| - ln|33/32|) = (2/5) * (ln(33) - ln(33) + ln(32)) = (2/5) * ln(32).Simplify Result: Simplify the result. The final answer is (2/5) * ln(32). Since 32 = 2^5, we can use the property of logarithms that ln(a^b) = b * ln(a) to simplify further: (2/5) * ln(2^5) = (2/5) * 5 * ln(2) = 2 * ln(2).

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$\int_{\frac{1}{2}}^{2}\frac{2x^{4}}{x^{5}+1}\,dx$

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Evaluate definite integrals using the chain rule

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\int_{\frac{1}{2}}^{2}\frac{2x^{4}}{x^{5}+1}\,dx

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $\frac{2x^4}{x^5 + 1}$ from $x = \frac{1}{2}$ to $x = 2$ .
Simplify Integral: Simplify the integral if possible. $\newline$ In this case, the integral does not simplify easily, so we proceed with the given integral.
Perform Substitution: Perform a substitution if applicable. $\newline$ Let $u = x^5 + 1$ , then $du = 5x^4 dx$ . We need to adjust the integral to match this substitution. To do this, we multiply and divide the integral by $5$ to get the $x^4$ term to match $du$ .
Rewrite in terms of $u$ : Rewrite the integral in terms of $u$ . $\newline$ The integral becomes $\frac{2}{5} \int_{\left(\frac{1}{2}\right)^2}^{2} \frac{5x^4}{x^5 + 1} \, dx = \frac{2}{5} \int_{u\left(\frac{1}{2}\right)}^{u(2)} \frac{du}{u}$ .
Update Limits: Update the limits of integration. $\newline$ When $x = \frac{1}{2}$ , $u = \left(\frac{1}{2}\right)^5 + 1 = \frac{1}{32} + 1 = \frac{33}{32}$ . $\newline$ When $x = 2$ , $u = 2^5 + 1 = 32 + 1 = 33$ .
Evaluate Integral: Evaluate the integral with the new limits. $\newline$ The integral is now $(\frac{2}{5}) \int_{\frac{33}{32}}^{33} \frac{du}{u}$ . The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|$ , so we have $(\frac{2}{5}) [\ln|u|]$ from $u = \frac{33}{32}$ to $u = 33$ .
Calculate Definite Integral: Calculate the definite integral. Plug in the limits of integration: $(\frac{2}{5}) \cdot (\ln|33| - \ln|\frac{33}{32}|) = (\frac{2}{5}) \cdot (\ln(33) - \ln(33) + \ln(32)) = (\frac{2}{5}) \cdot \ln(32)$ .
Simplify Result: Simplify the result. $\newline$ The final answer is $(\frac{2}{5}) \cdot \ln(32)$ . Since $32 = 2^5$ , we can use the property of logarithms that $\ln(a^b) = b \cdot \ln(a)$ to simplify further: $(\frac{2}{5}) \cdot \ln(2^5) = (\frac{2}{5}) \cdot 5 \cdot \ln(2) = 2 \cdot \ln(2)$ .