Sava's question via email about integration with partial fractions.

In summary, the integral can be evaluated using partial fraction decomposition and the answer is $-3\ln{\left| x - 1 \right| } + \frac{1}{x - 1} + 4\ln{\left| x - 2 \right| } + C$.
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Evaluate $\displaystyle \begin{align*} \int{\frac{x^2}{\left( x - 1 \right) ^2 \, \left( x - 2 \right) } \,\mathrm{d}x} \end{align*}$

As there is a repeated root, the partial fraction decomposition we should use is:

$\displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right) ^2 } + \frac{C}{x - 2} &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ \frac{A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 }{\left( x -1 \right) ^2 \,\left( x - 2 \right) } &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 1 \end{align*}$ to find $\displaystyle \begin{align*} -B = 1 \implies B = -1 \end{align*}$

Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} C = 4 \end{align*}$

Substitute B and C back in:

$\displaystyle \begin{align*} A\,\left( x - 1 \right) \left( x - 2 \right) - \left( x - 2 \right) + 4\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 0 \end{align*}$ to find $\displaystyle \begin{align*} 2\,A + 2 + 4 = 0 \implies 2\,A = -6 \implies A = -3 \end{align*}$.

So that means

$\displaystyle \begin{align*} \int{ \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \,\mathrm{d}x} &= \int{ \left[ -\frac{3}{x - 1} - \frac{1}{\left( x - 1 \right) ^2 } + \frac{4}{x - 2} \right] \,\mathrm{d}x } \\ &= -3\ln{\left| x - 1 \right| } + \frac{1}{x - 1} + 4\ln{\left| x - 2 \right| } + C \end{align*}$
 
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As there is a repeated root, the partial fraction decomposition we should use is:

$\displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right) ^2 } + \frac{C}{x - 2} &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ \frac{A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 }{\left( x -1 \right) ^2 \,\left( x - 2 \right) } &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 1 \end{align*}$ to find $\displaystyle \begin{align*} -B = 1 \implies B = -1 \end{align*}$

Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} C = 4 \end{align*}$

Substitute B and C back in:

$\displaystyle \begin{align*} A\,\left( x - 1 \right) \left( x - 2 \right) - \left( x - 2 \right) + 4\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 0 \end{align*}$ to find $\displaystyle \begin{align*} 2\,A + 2 + 4 = 0 \implies 2\,A = -6 \implies A = -3 \end{align*}$.

So that means

$\displaystyle \begin{align*} \int{ \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \,\mathrm{d}x} &= \int{ \left[ -\frac{3}{x - 1} - \frac{1}{\left( x - 1 \right) ^2 } + \frac{4}{x - 2} \right] \,\mathrm{d}x } \\ &= -3\ln{\left| x - 1 \right| } + \frac{1}{x - 1} + 4\ln{\left| x - 2 \right| } + C \end{align*}$
This is correct!
 
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FAQ: Sava's question via email about integration with partial fractions.

What is integration with partial fractions?

Integration with partial fractions is a method used in calculus to simplify and solve integrals with rational functions. It involves breaking down a complex rational function into simpler fractions, which can then be integrated more easily.

When is integration with partial fractions used?

This method is commonly used when integrating rational functions that cannot be integrated through other methods such as substitution or integration by parts. It is also used in cases where the degree of the numerator is equal to or greater than the degree of the denominator.

What are the steps for integration with partial fractions?

The first step is to factor the denominator of the rational function. Then, use the identified factors to write the rational function as a sum of simpler fractions. Next, determine the unknown coefficients by setting up and solving a system of equations. Finally, integrate each individual fraction and combine the results to get the final solution.

What is the importance of integration with partial fractions in mathematics?

Integration with partial fractions is an important tool in calculus, as it allows for the solution of integrals that would otherwise be difficult or impossible to solve. It is also used in various areas of mathematics, such as in differential equations and in the Laplace transform.

Are there any applications of integration with partial fractions in real life?

Yes, integration with partial fractions has various applications in real life, such as in engineering, physics, and economics. It is used to solve problems related to optimization, motion, and growth, among others. It is also used in the analysis of circuits and in the calculation of probabilities in statistics.

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