Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{2+x^{2}}{1+x^{2}} d x$$

Short answer

Question: Evaluate the indefinite integral of the given function: $$\int \frac{2+x^2}{1+x^2} dx$$ Answer: The indefinite integral of the given function is: $$(x + \arctan(x)) + C$$

Step-by-step solution

Rewrite the integrand as a sum

To simplify the integrand, let's write it as the sum of two terms: $$\int \frac{2+x^{2}}{1+x^{2}} dx = \int \left( \frac{2}{1+x^{2}} + \frac{x^2}{1+x^2} \right) dx$$

Integrate each term separately

Notice that the second term is easy to integrate as it is already in a simple form. The first term can be recognized as the derivative of \(\arctan(x)\). Now, integrate each term separately: $$\int \frac{2}{1+x^{2}} dx = 2\int \frac{1}{1+x^{2}} dx = 2\arctan(x) + C_1$$ $$\int \frac{x^2}{1+x^2} dx = \int \left( 1 - \frac{1}{1+x^2} \right) dx = x - \arctan(x) + C_2$$

Combine the results and find the general solution

Now, we can combine the results to get the general solution: $$\int \frac{2+x^2}{1+x^2} dx = 2\arctan(x) + x - \arctan(x) + C$$ Where \(C = C_1 + C_2\) is a constant representing the constant of integration. So, the indefinite integral is: $$\int \frac{2+x^2}{1+x^2} dx = (x + \arctan(x)) + C$$

Check the result by differentiation

To verify our result, let's find the derivative of our solution: $$\frac{d}{dx} (x + \arctan(x)) = 1 + \frac{1}{1 + x^2}$$ Notice that this expression is slightly different from the integrand. Therefore, we need to further simplify the integrand: $$\frac{2+x^2}{1+x^2} = \frac{1}{1+x^2} + \frac{x^2 + 1}{1+x^2} = \frac{1}{1+x^2} + 1.$$ Now, comparing this expression with our derivative, we can see they match, confirming that the integral evaluation is correct.