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int_(1)^(sqrt5)(2ln(x^(2)+7))/(x^(2)+2)dx

$\int_{1}^{\sqrt{5}} \frac{2 \ln \left(x^{2}+7\right)}{x^{2}+2} d x$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{1}^{\sqrt{5}} \frac{2 \ln \left(x^{2}+7\right)}{x^{2}+2} d x

Substitution: Let's do a substitution: let $u = x^2 + 7$ . Then $du = 2x \, dx$ .
Change Limits: Change the limits of integration. When $x = 1$ , $u = 1^2 + 7 = 8$ . When $x = \sqrt{5}$ , $u = (\sqrt{5})^2 + 7 = 12$ .
Rewrite Integral: Rewrite the integral in terms of $u$ : $\int_{8}^{12}\frac{\ln(u)}{u-2} \, du$ .
Integrate $\ln(u)/(u-2)$ : Now, we need to integrate $\ln(u)/(u-2)$ . This isn't a standard integral, so we might need to use integration by parts.
Integration by Parts: Let's use integration by parts: let $v = \ln(u)$ and $dw = \frac{1}{(u-2)} du$ . Then $dv = \frac{1}{u} du$ and $w = \ln(u-2)$ .
Apply Integration by Parts: Apply integration by parts: $\int v \, dw = vw - \int w \, dv$ .
Calculate Integral: Calculate the integral: $\ln(u) \cdot \ln(u-2) - \int \ln(u-2) \cdot \left(\frac{1}{u}\right) du$ .
Correct Mistake: This step is incorrect because the integral of $\frac{1}{(u-2)}$ is not $\ln(u-2)$ . The correct antiderivative should be a logarithm with a constant adjustment. We need to correct this mistake.

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