int_(1)^(2)((3)/(x^(2))-1)dx

Identify integral: Identify the integral to be solved. We need to evaluate the integral of the function (3/x^2) - 1 with respect to x from 1 to 2.Break into two integrals: Break the integral into two separate integrals. The integral of a sum or difference of functions is the sum or difference of their integrals. Therefore, we can write: ∫(3/x^2 - 1)dx = ∫(3/x^2)dx - ∫1dxEvaluate first integral: Evaluate the first integral ∫(3/x^2)dx. The integral of 3/x^2 with respect to x is -3/x. This is because the integral of x^n is x^(n+1)/(n+1) for n ≠ -1, and in this case, n = -2. So, ∫(3/x^2)dx = -3/x + CEvaluate second integral: Evaluate the second integral ∫1dx. The integral of 1 with respect to x is x, because the derivative of x with respect to x is 1. So, ∫1dx = x + CCombine results: Combine the results of Step 3 and Step 4. The combined integral from 1 to 2 of (3/x^2) - 1 is: ∫(3/x^2 - 1)dx = (-3/x + x) from 1 to 2Evaluate definite integral: Evaluate the definite integral from 1 to 2. We substitute the upper limit of integration (x = 2) and then the lower limit (x = 1) into the antiderivative and subtract the two results: (-3/2 + 2) - (-3/1 + 1) = (-1.5 + 2) - (-3 + 1) = 0.5 - (-2) = 0.5 + 2 = 2.5

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$15$ . $\int_{1}^{2}\left(\frac{3}{x^{2}}-1\right) d x$

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Evaluate definite integrals using the chain rule

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15

\int_{1}^{2}\left(\frac{3}{x^{2}}-1\right) d x

Identify integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $\frac{3}{x^2} - 1$ with respect to $x$ from $1$ to $2$ .
Break into two integrals: Break the integral into two separate integrals. $\newline$ The integral of a sum or difference of functions is the sum or difference of their integrals. Therefore, we can write: $\newline$ $\int(\frac{3}{x^2} - 1)\,dx = \int(\frac{3}{x^2})\,dx - \int 1\,dx$
Evaluate first integral: Evaluate the first integral $\int(\frac{3}{x^2})dx$ . The integral of $\frac{3}{x^2}$ with respect to $x$ is $-\frac{3}{x}$ . This is because the integral of $x^n$ is $\frac{x^{n+1}}{n+1}$ for $n \neq -1$ , and in this case, $n = -2$ . So, $\int(\frac{3}{x^2})dx = -\frac{3}{x} + C$
Evaluate second integral: Evaluate the second integral $\int 1 \, dx$ . The integral of $1$ with respect to $x$ is $x$ , because the derivative of $x$ with respect to $x$ is $1$ . So, $\int 1 \, dx = x + C$
Combine results: Combine the results of Step $3$ and Step $4$ . $\newline$ The combined integral from $1$ to $2$ of $(3/x^2) - 1$ is: $\newline$ $\int (3/x^2 - 1)dx = (-3/x + x)$ from $1$ to $2$
Evaluate definite integral: Evaluate the definite integral from $1$ to $2$ . We substitute the upper limit of integration ( $x = 2$ ) and then the lower limit ( $x = 1$ ) into the antiderivative and subtract the two results: $(-\frac{3}{2} + 2) - (-\frac{3}{1} + 1) = (-1.5 + 2) - (-3 + 1) = 0.5 - (-2) = 0.5 + 2 = 2.5$