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Chapter 17: Problem 19

Find \(\int \frac{1}{x^{2}-4 x+3} \mathrm{~d} x\).

Short Answer

Expert verified
Question: Find the integral of the following function: \(\int \frac{1}{x^{2}-4 x+3} \mathrm{~d} x\) Answer: The integral of the given function is \(\int \frac{1}{x^{2}-4 x+3} \mathrm{~d} x = \frac{1}{2} \ln\left|\frac{x-3}{x-1}\right| + C\).

Step by step solution

01

Factor the denominator

Factor the expression in the denominator as follows: \((x^{2}-4x+3) = (x-3)(x-1)\) So, the function becomes \(\int \frac{1}{(x-3)(x-1)} \mathrm{~d} x\).
02

Perform Partial Fraction Decomposition

We use partial fraction decomposition to rewrite the integrand as a sum of two simpler fractions: \(\frac{1}{(x-3)(x-1)} = \frac{A}{x-3} + \frac{B}{x-1}\), where A and B are constants. Now, we find the constants A and B: \(1 = A(x-1) + B(x-3)\) To find A, let \(x=3\): \(1 = A(2) \Rightarrow A=\frac{1}{2}\) To find B, let \(x=1\): \(1 = B(-2) \Rightarrow B=-\frac{1}{2}\) Now, rewrite the integrand as the sum of these simpler fractions: \(\int \frac{1}{(x-1)(x-3)} \mathrm{~d} x = \int \left[ \frac{1/2}{x-3} - \frac{1/2}{x-1} \right] \mathrm{~d} x\)
03

Find the Antiderivative of Each Term

Now, we find the antiderivative of each term in the sum: \(\int \left[ \frac{1/2}{x-3} - \frac{1/2}{x-1} \right] \mathrm{~d} x = \frac{1}{2}\int \frac{1}{x-3} \mathrm{~d} x - \frac{1}{2}\int \frac{1}{x-1} \mathrm{~d} x\) Both integrals are now simple logarithmic integrals: \(= \frac{1}{2} \ln|x-3| - \frac{1}{2} \ln|x-1| + C\)
04

Combine the Logarithmic Terms (if needed)

In this case, we can leave the answer as it is. However, sometimes it is beneficial to combine logarithmic terms using the logarithmic identity \(\ln a - \ln b = \ln \frac{a}{b}\): \(= \frac{1}{2} \ln\left|\frac{x-3}{x-1}\right| + C\) The integral of the given function is: \(\int \frac{1}{x^{2}-4 x+3} \mathrm{~d} x = \frac{1}{2} \ln\left|\frac{x-3}{x-1}\right| + C\).

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