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Chapter 4: Problem 108

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

Short Answer

Expert verified
$$\int \frac{2+x^{2}}{1+x^{2}} d x$$

Step by step solution

01

Decompose the integrand

Decompose the fraction in this way: $$\frac{2+x^2}{1+x^2} = \frac{1+x^2}{1+x^2} - \frac{1}{1+x^2} = 1 - \frac{1}{1+x^2}$$ Now our integral becomes: $$\int \left(1-\frac{1}{1+x^2}\right) dx $$
02

Split the integral

Now, let's split our integral into two separate integrals: $$\int \left(1-\frac{1}{1+x^2}\right) dx = \int 1 dx - \int \frac{1}{1+x^2} dx$$
03

Integrate both parts

Now, we'll integrate both parts separately. For the first integral: $$\int 1 dx = x + C_1$$ where \(C_1\) is a constant of integration. For the second integral, recall the following integral formula: $$\int \frac{1}{1+x^2}dx = \arctan(x) + C_2$$ where \(C_2\) is another constant of integration. Thus, our integrals become: $$x + C_1 - \arctan(x) - C_2$$
04

Combine the constants and write the final answer

Combine the constants \(C_1\) and \(C_2\) into a single constant \(C\). The final answer for the indefinite integral is: $$x - \arctan(x) + C$$
05

Check the result by differentiating

Differentiate our result with respect to \(x\) to check if it matches the original function: $$\frac{d}{dx}(x - \arctan(x) + C) = 1 - \frac{d}{dx}(\arctan(x)) = 1 - \frac{1}{1+x^2}$$ As our derived function matches the original function, it confirms that our result was correct.

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Most popular questions from this chapter

Suppose \(f(x)=1 /(1+x)\) is to be approximated near \(x=0\). Find the linear approximation to \(f\) at 0 . Then complete the following table showing the errors in various approximations. Use a calculator to obtain the exact values. The percent error is \(100 \cdot |\) approximation \(-\) exact \(|/|\) exact \(| .\) Comment on the behavior of the errors as \(x\) approaches 0 .

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