int(3x-5)/((x-2)^(2))dx

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of the function (3x-5)/((x-2)^(2)) with respect to x.Decompose Integrand: Decompose the integrand into partial fractions if possible. However, in this case, the numerator is a linear expression and the denominator is a square of a linear expression, which suggests that we should try a different approach, such as looking for a direct antiderivative or using substitution.Use Substitution: Use substitution to simplify the integral. Let u = x - 2, then du = dx. The integral becomes: ∫(3(u+2)-5)/u^2 du Simplify the integrand: ∫(3u+6-5)/u^2 du ∫(3u+1)/u^2 duSplit Integral: Split the integral into two simpler integrals. ∫(3u/u^2) du + ∫(1/u^2) du ∫3/u du - ∫1/u^2 duIntegrate Terms: Integrate each term separately. The integral of 3/u with respect to u is 3ln|u|, and the integral of 1/u^2 with respect to u is -1/u. So we have: 3ln|u| - 1/u + C, where C is the constant of integration.Substitute Back: Substitute back the original variable x into the antiderivative. Since u = x - 2, we have: 3ln|x-2| - 1/(x-2) + C

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Calculus

Evaluate definite integrals using the chain rule

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\int \frac{3 x-5}{(x-2)^{2}} d x

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $(3x-5)/((x-2)^{2})$ with respect to $x$ .
Decompose Integrand: Decompose the integrand into partial fractions if possible. However, in this case, the numerator is a linear expression and the denominator is a square of a linear expression, which suggests that we should try a different approach, such as looking for a direct antiderivative or using substitution.
Use Substitution: Use substitution to simplify the integral. $\newline$ Let $u = x - 2$ , then $du = dx$ . The integral becomes: $\newline$ $\int\frac{3(u+2)-5}{u^2} du$ $\newline$ Simplify the integrand: $\newline$ $\int\frac{3u+6-5}{u^2} du$ $\newline$ $\int\frac{3u+1}{u^2} du$
Split Integral: Split the integral into two simpler integrals. $\newline$ $\int(\frac{3u}{u^2}) du + \int(\frac{1}{u^2}) du$ $\newline$ $\int\frac{3}{u} du - \int\frac{1}{u^2} du$
Integrate Terms: Integrate each term separately. $\newline$ The integral of $3/u$ with respect to $u$ is $3\ln|u|$ , and the integral of $1/u^2$ with respect to $u$ is $-1/u$ . $\newline$ So we have: $\newline$ $3\ln|u| - 1/u + C$ , where $C$ is the constant of integration.
Substitute Back: Substitute back the original variable $x$ into the antiderivative. $\newline$ Since $u = x - 2$ , we have: $\newline$ $3\ln|x-2| - \frac{1}{x-2} + C$