Problem 21 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The final result of the integral is \(-\ln|x^{-1}+1|+C\).

Step 1: Perform substitution

Let's perform a substitution to simplify the integrand, to do this, let \(u = x^{-1}+1\). Then, we have to find \(dx\) in terms of \(du\). First, let's find \(\frac{du}{dx}\): $$\frac{du}{dx} = \frac{d}{dx} (x^{-1} + 1) = -x^{-2}$$ Now, we can find \(dx\) in terms of \(du\): $$dx = \frac{du}{-x^{-2}}$$ Now we can rewrite the integral in terms of u: $$\int \frac{1}{u}\frac{du}{-x^{-2}}$$

Step 2: Simplify the integral

Now we can simplify the integral. Since we know that \(u = x^{-1}+1\), we can substitute \(u\) back in terms of x: $$\int \frac{-1}{x^{-2}(x^{-1}+1)} du$$ Now we can perform the integral in terms of u: $$\int \frac{-1}{x^{-2}(x^{-1}+1)} du = -\int \frac{1}{u} du$$

Step 3: Integrate

The integral of \(\frac{1}{u}\) is the natural logarithm, so we have: $$-\int \frac{1}{u} du = -\ln|u|+C$$

Step 4: Rewrite the result in terms of x

Now, we can rewrite the final result back in terms of x, using our initial substitution \(u = x^{-1}+1\): $$-\ln|u|+C = -\ln|x^{-1}+1|+C$$ Thus, the original integral can be written as: $$\int \frac{dx}{x^{-1}+1} = -\ln|x^{-1}+1|+C$$