int(x-3)/((x-1)(x-2))dx

Set up decomposition: We need to integrate the function (x-3)/((x-1)(x-2)). To do this, we will use partial fraction decomposition to express the integrand as a sum of simpler fractions that we can integrate individually.Find A and B: First, we set up the partial fraction decomposition: (x-3)/((x-1)(x-2)) = A/(x-1) + B/(x-2) We need to find the values of A and B.Plug in x values: To find A and B, we multiply both sides of the equation by (x-1)(x-2) to get rid of the denominators: (x-3) = A(x-2) + B(x-1)Calculate integrals: Now we will find the values of A and B by plugging in convenient values for x. Let's first plug in x = 1: (1-3) = A(1-2) + B(1-1) -2 = -A A = 2Integrate 2/(x-1): Next, we plug in x = 2: (2-3) = A(2-2) + B(2-1) -1 = B B = -1Integrate 1/(x-2): Now that we have A and B, we rewrite the original integral: ∫(x-3)/((x-1)(x-2))dx = ∫(2/(x-1) - 1/(x-2))dxCombine results: We can now integrate each term separately: ∫(2/(x-1))dx - ∫(1/(x-2))dxFinal indefinite integral: The integral of 2/(x-1) is 2ln|x-1|, and the integral of 1/(x-2) is ln|x-2|. So we have: 2ln|x-1| - ln|x-2| + C, where C is the constant of integration.We have found the indefinite integral of the given function: ∫(x-3)/((x-1)(x-2))dx = 2ln|x-1| - ln|x-2| + C

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\int \frac{x-3}{(x-1)(x-2)} d x

Set up decomposition: We need to integrate the function $(x-3)/((x-1)(x-2))$ . To do this, we will use partial fractions" target="_blank" class="backlink">fraction decomposition to express the integrand as a sum of simpler fractions that we can integrate individually.
Find $A$ and $B$ : First, we set up the partial fraction decomposition: $\newline$ $(x-3)/((x-1)(x-2)) = \frac{A}{(x-1)} + \frac{B}{(x-2)}$ $\newline$ We need to find the values of $A$ and $B$ .
Plug in x values: To find $A$ and $B$ , we multiply both sides of the equation by $(x-1)(x-2)$ to get rid of the denominators: $\newline$ $(x-3) = A(x-2) + B(x-1)$
Calculate integrals: Now we will find the values of $A$ and $B$ by plugging in convenient values for $x$ . Let's first plug in $x = 1$ :
$(1-3) = A(1-2) + B(1-1)$
$-2 = -A$
$A = 2$
Integrate $\frac{2}{x-1}$ : Next, we plug in $x = 2$ :
$(2-3) = A(2-2) + B(2-1)$
$-1 = B$
$B = -1$
Integrate $\frac{1}{x-2}$ : Now that we have $A$ and $B$ , we rewrite the original integral: $\newline$ \int\frac{x$-3$}{(x$-1$)(x$-2$)}dx = \int\left(\frac{$2$}{x$-1$} - \frac{$1$}{x$-2$}\right)dx
Combine results: We can now integrate each term separately: $\int\left(\frac{\(2$}{x$-1$}\right)dx - \int\left(\frac{$1$}{x$-2$}\right)dx
Final indefinite integral: The integral of \(\frac{2}{x-1} is $2\ln|x-1|$ , and the integral of $\frac{1}{x-2}$ is $\ln|x-2|$ . So we have: $\newline$ $2\ln|x-1| - \ln|x-2| + C$ , where $C$ is the constant of integration.
Final indefinite integral: The integral of $\frac{2}{x-1}$ is $2\ln|x-1|$ , and the integral of $\frac{1}{x-2}$ is $\ln|x-2|$ . So we have: $\newline$ $2\ln|x-1| - \ln|x-2| + C$ , where $C$ is the constant of integration.We have found the indefinite integral of the given function: $\newline$ $\int\frac{x-3}{(x-1)(x-2)}dx = 2\ln|x-1| - \ln|x-2| + C$