(g) int_(1)^(2)(1+2ln(x))/((1+ln(x))^(2)x)dx

Identify Substitution: Let's start by identifying a substitution that will simplify the integral. We can let u = 1 + ln(x), which implies that du = (1/x) dx. This substitution will simplify the integral significantly.Find Limits for u: Now, we need to find the corresponding limits of integration for u. When x = 1, u = 1 + ln(1) = 1. When x = 2, u = 1 + ln(2). So, the new limits of integration are from u = 1 to u = 1 + ln(2).Substitute u into Integral: Substitute u = 1 + ln(x) and du = (1/x) dx into the integral. The integral becomes: ∫(1+2ln(x))/((1+ln(x))^2*x) dx = ∫(1+2(u-1))/u^2 duSimplify Integral: Simplify the integral: ∫(1+2(u-1))/u^2 du = ∫(1+2u-2)/u^2 du = ∫(1/u^2 + 2u/u^2 - 2/u^2) du = ∫(1/u^2 + 2/u - 2/u^2) du = ∫(2/u - 1/u^2) duIntegrate Term by Term: Now, integrate term by term: ∫(2/u - 1/u^2) du = 2∫(1/u) du - ∫(1/u^2) du = 2ln|u| - (1/u) + CSubstitute back to x: Substitute back u = 1 + ln(x) to get the antiderivative in terms of x: 2ln|u| - (1/u) + C = 2ln|1 + ln(x)| - 1/(1 + ln(x)) + CEvaluate Definite Integral: Evaluate the definite integral from u = 1 to u = 1 + ln(2): [2ln|1 + ln(x)| - 1/(1 + ln(x))] from x = 1 to x = 2 = (2ln(1 + ln(2)) - 1/(1 + ln(2))) - (2ln(1) - 1/1) = (2ln(1 + ln(2)) - 1/(1 + ln(2))) - (0 - 1) = 2ln(1 + ln(2)) - 1/(1 + ln(2)) + 1

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$\int_{1}^{2}\frac{1+2\ln(x)}{(1+\ln(x))^{2}x}\,dx$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{1}^{2}\frac{1+2\ln(x)}{(1+\ln(x))^{2}x}\,dx

Identify Substitution: Let's start by identifying a substitution that will simplify the integral. We can let $u = 1 + \ln(x)$ , which implies that $du = (\frac{1}{x}) dx$ . This substitution will simplify the integral significantly.
Find Limits for $u$ : Now, we need to find the corresponding limits of integration for $u$ . When $x = 1$ , $u = 1 + \ln(1) = 1$ . When $x = 2$ , $u = 1 + \ln(2)$ . So, the new limits of integration are from $u = 1$ to $u = 1 + \ln(2)$ .
Substitute $u$ into Integral: Substitute $u = 1 + \ln(x)$ and $du = (\frac{1}{x}) dx$ into the integral. The integral becomes: $\newline$ $\int \frac{1+2\ln(x)}{(1+\ln(x))^2\cdot x} dx = \int \frac{1+2(u-1)}{u^2} du$
Simplify Integral: Simplify the integral: \int\left(\frac{ $1$ + $2$ (u $-1$ )}{u^ $2$ }\right) du = \int\left(\frac{ $1$ + $2$ u $-2$ }{u^ $2$ }\right) du = \int\left(\frac{ $1$ }{u^ $2$ } + \frac{ $2$ u}{u^ $2$ } - \frac{ $2$ }{u^ $2$ }\right) du = \int\left(\frac{ $1$ }{u^ $2$ } + \frac{ $2$ }{u} - \frac{ $2$ }{u^ $2$ }\right) du = \int\left(\frac{ $2$ }{u} - \frac{ $1$ }{u^ $2$ }\right) du
Integrate Term by Term: Now, integrate term by term: $\int(\frac{2}{u} - \frac{1}{u^2}) du = 2\int(\frac{1}{u}) du - \int(\frac{1}{u^2}) du = 2\ln|u| - (\frac{1}{u}) + C$
Substitute back to x: Substitute back $u = 1 + \ln(x)$ to get the antiderivative in terms of x: $\newline$ $2\ln|u| - (1/u) + C = 2\ln|1 + \ln(x)| - 1/(1 + \ln(x)) + C$
Evaluate Definite Integral: Evaluate the definite integral from $u = 1$ to $u = 1 + \ln(2)$ : $\left[2\ln|1 + \ln(x)| - \frac{1}{1 + \ln(x)}\right]_{x = 1}^{x = 2}$ $= (2\ln(1 + \ln(2)) - \frac{1}{1 + \ln(2)}) - (2\ln(1) - \frac{1}{1})$ $= (2\ln(1 + \ln(2)) - \frac{1}{1 + \ln(2)}) - (0 - 1)$ $= 2\ln(1 + \ln(2)) - \frac{1}{1 + \ln(2))} + 1$