Question

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{1}^{3} \frac{2}{x^{2}+2 x+1} d x$$

Short answer

Question: Evaluate the definite integral: $\int_{1}^{3} \frac{2}{x^{2}+2 x+1} d x$ Answer: $-\frac{1}{2}$

Step-by-step solution

Factor the denominator

The first thing we want to do is see if it is possible to factor the denominator or simplify the function inside the integral. To do this, consider the quadratic inside the integral: $$ x^2 + 2x + 1 = (x + 1)^2 $$ The denominator can be factored. So, we can rewrite the integral as follows: $$ \int_{1}^{3} \frac{2}{(x + 1)^2} dx $$

Substitute

Now, let's use the substitution method to simplify the integration. Let: $$ u = x + 1 \Rightarrow \frac{du}{dx} = 1 \Rightarrow du = dx $$ With this substitution, we also have to change our limits of integration. When \(x = 1\), we have \(u = 2\), and when \(x = 3\), we have \(u = 4\). So, our integral becomes: $$ \int_{2}^{4} \frac{2}{u^2} du $$

Integrate

Now, we can integrate the new function with respect to u: $$ \int \frac{2}{u^2} du = -\frac{2}{u} + C $$ After finding the antiderivative, all that remains is to substitute our values for u and calculate the definite integral.

Evaluate the definite integral

Replace u with x + 1 and apply the limits of integration: $$ -\frac{2}{u}|_{2}^{4} = \left(-\frac{2}{x+1}\right) \Big|_{x=1}^{x=3} $$ $$ = \left(-\frac{2}{4}\right) - \left(-\frac{2}{2}\right) = \frac{1}{2} - 1 = -\frac{1}{2} $$ The integral evaluates to -1/2. The answer to the given exercise is: $$ \int_{1}^{3} \frac{2}{x^{2}+2 x+1} d x = -\frac{1}{2} $$