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Chapter 6: Problem 13

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{3} \frac{2 x-1}{x+1} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{0}^{3} \frac{2 x-1}{x+1} d x$$. Answer: The value of the definite integral is $$4\ln4$$.

Step by step solution

01

Finding the antiderivative of the function

To find the antiderivative, we can use the substitution method. Let's substitute $$u = x + 1$$, which means $$du = dx$$. Now rewrite the integral in terms of $$u$$: $$\int \frac{2(x-1)}{u} d u$$ Now, let's find the antiderivative: $$\int \frac{2(x-1)}{u} d u = 2(x-1)\ln|u| + C$$ Now let's substitute back $$u = x + 1$$: $$F(x) = 2(x-1)\ln|x + 1| + C$$ This is the antiderivative of the function.
02

Evaluate the antiderivative at the limits of integration

Now we need to evaluate $$F(x)$$ at the limits of integration, which are 3 and 0. To do this, we plug in the limits and calculate their values: $$F(3) = 2(3-1)\ln|3+1| = 4\ln4$$ $$F(0) = 2(0-1)\ln|0+1| = -2\ln1$$
03

Subtract the value of the antiderivative at 0 from the value at 3

Now we need to subtract the value of $$F(0)$$ from $$F(3)$$ to get the value of the definite integral: $$\int_{0}^{3} \frac{2 x-1}{x+1} d x = F(3) - F(0) = 4\ln4 - (-2\ln1)$$ Since $$\ln1 = 0$$, the expression simplifies to: $$\int_{0}^{3} \frac{2 x-1}{x+1} d x = 4\ln4$$ So, the value of the definite integral is $$4\ln4$$.

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