Problem 30 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x$$. Answer: $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x = 8 + 3\ln |3|$$

Step 1: Division

Use polynomial long division or synthetic division to separate the terms. For the first term, divide \(x^2\) by \(x-1\). For the second term, divide 2 by \(x-1\). After performing the division, we get \(\frac{x^2}{x-1}=x+1+\frac{1}{x-1}\) and \(\frac{2}{x-1}=\frac{2}{x-1}\). So, the integral can be written as: $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x=\int_{2}^{4} (x+1+\frac{1}{x-1}+\frac{2}{x-1}) d x$$

Step 2: Separate the integrals

Now we have: $$\int_{2}^{4} (x+1+\frac{1}{x-1}+\frac{2}{x-1}) d x = \int_{2}^{4} x d x + \int_{2}^{4} 1 d x + \int_{2}^{4} \frac{1}{x-1} d x + \int_{2}^{4} \frac{2}{x-1} d x$$

Step 3: Integrate the separated terms

Now integrate each term. $$\int_{2}^{4} x d x = \frac{x^2}{2}\Big|_{2}^{4}$$ $$\int_{2}^{4} 1 d x = x\Big|_{2}^{4}$$ $$\int_{2}^{4} \frac{1}{x-1} d x = \ln |x-1|\Big|_{2}^{4}$$ $$\int_{2}^{4} \frac{2}{x-1} d x = 2 \ln |x-1|\Big|_{2}^{4}$$

Step 4: Evaluate the definite integration

Now we can just plug in the limits of integration and subtract: $$\frac{x^2}{2}\Big|_{2}^{4}=8-2=6$$ $$x\Big|_{2}^{4}=4-2=2$$ $$\ln |x-1|\Big|_{2}^{4}=\ln |3|-\ln |1|=\ln |3|$$ $$2 \ln |x-1|\Big|_{2}^{4}=2(\ln |3|-\ln |1|)=2 \ln |3|$$

Step 5: Combine the results

Now, sum up the results obtained from the previous step: $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x = 6 + 2 + \ln |3| + 2\ln |3|= 8 + 3\ln |3|$$ So the value of the integral is: $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x = 8 + 3\ln |3|$$