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Chapter 7: Problem 35

Evaluate the following integrals. $$\int \frac{x^{2}-x}{(x-2)(x-3)^{2}} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the given function: $$\int \frac{x^{2}-x}{(x-2)(x-3)^{2}} dx$$ Answer: $$\int \frac{x^{2}-x}{(x-2)(x-3)^{2}} d x = 2 \ln |x-2| - 3 \ln |x-3| - \frac{6}{x-3} + C$$

Step by step solution

01

Decompose the fraction into partial fractions

To find the partial fraction decomposition, we first write the fraction as a sum of partial fractions: $$\frac{x^{2}-x}{(x-2)(x-3)^{2}} = \frac{A}{x-2} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$ Multiply both sides of the equation by the common denominator \((x-2)(x-3)^2\) to simplify: $$x^2 - x = A(x-3)^2 + B(x-2)(x-3) + C(x-2)$$
02

Solve for A, B, and C

To find the values of A, B, and C, we can solve this equation by plugging in suitable values for x. First, let x = 2: $$2^2 - 2 = A(2-3)^2 \Rightarrow A = 2$$ Next, let x = 3: $$3^2 - 3 = C(3-2) \Rightarrow C = 6$$ Finally, we differentiate the equation with respect to x and plug in x = 3 to solve for B: $$2x - 1 = 2(2)(x-3) + B(2x-5) + 6$$ $$2(3) -1 = 2(2)(3-3) + B(2\cdot3-5) + 6$$ $$B = -3$$ Now the partial fraction decomposition is: $$\frac{x^{2}-x}{(x-2)(x-3)^{2}} = \frac{2}{x-2} - \frac{3}{x-3} + \frac{6}{(x-3)^2}$$
03

Integrate each term

Now, we integrate each term of the sum with respect to x: $$\int \frac{x^{2}-x}{(x-2)(x-3)^{2}} d x = \int \frac{2}{x-2} dx - \int \frac{3}{x-3} dx + \int \frac{6}{(x-3)^2} dx$$
04

Evaluate the integrals

We can find the antiderivatives for each integral using the power rule and substitution method: $$2 \int \frac{1}{x-2} dx - 3 \int \frac{1}{x-3} dx + 6 \int \frac{1}{(x-3)^2} dx = 2 \ln |x-2| - 3 \ln |x-3| - 6 \cdot \frac{1}{x-3}$$
05

Combine the antiderivatives and add the constant of integration

The final answer will be the sum of the antiderivatives plus a constant of integration, denoted by C: $$\int \frac{x^{2}-x}{(x-2)(x-3)^{2}} d x = 2 \ln |x-2| - 3 \ln |x-3| - \frac{6}{x-3} + C$$

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