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Chapter 7: Problem 36

Evaluate the following integrals. $$\int \frac{x}{x^{4}+2 x^{2}+1} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the following function: $$\int\frac{x}{x^4+2x^2+1}dx$$ Answer: $$-\frac{1}{2}\frac{1}{x^2+1}+C$$

Step by step solution

01

Factor the denominator

The denominator is a quadratic in terms of \(x^2\), so let's factor it: $$x^4 + 2x^2 + 1 = (x^2 + 1)(x^2 + 1)$$ Now, our integral becomes: $$\int \frac{x}{(x^2 + 1)(x^2 + 1)} dx$$
02

Perform substitution

Let's make a substitution to simplify the integral: Let \(u = x^2 + 1 \implies du = 2x \ dx\). Now, solving for \(dx\), we get: $$dx = \frac{du}{2x}$$ Substitute \(u\) and \(dx\) in the original integral: $$\int \frac{x}{u^2} \cdot \frac{du}{2x}$$ Now, we can simplify and cancel the \(x\) in the numerator: $$\frac{1}{2} \int \frac{1}{u^2} du$$
03

Evaluate the integral

Now, we can evaluate the simplified integral: $$ \frac{1}{2} \int \frac{1}{u^2} du = \frac{1}{2} \int u^{-2} du$$ Applying the power rule for integration, we get: $$\frac{1}{2} \left[ \frac{u^{-1}}{-1} \right] + C = -\frac{1}{2} u^{-1} + C = -\frac{1}{2} \frac{1}{u} + C$$
04

Substitute back in terms of x

Now, substitute back \(u = x^2 + 1\): $$-\frac{1}{2} \frac{1}{x^2 + 1} + C$$ So, the evaluated integral is: $$\int \frac{x}{x^4 + 2x^2 + 1} dx = -\frac{1}{2} \frac{1}{x^2 + 1} + C$$

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