Problem 36 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the integral of the given function: $$\int \frac{x}{x^{4}+2 x^{2}+1} d x$$ Answer: The integral evaluated is: $$\int \frac{x}{x^{4}+2 x^{2}+1} d x = -\frac{1}{2(x^2+1)} + C$$

Step 1: Make a substitution

Let's make a substitution to simplify the integrand. Let \(u = x^2\). Then, differentiate \(u\) with respect to \(x\), we get: $$du = 2x \,dx$$ Now let's substitute \(u\) and \(du\) into the integral: $$\int \frac{x}{x^{4}+2 x^{2}+1} d x = \frac{1}{2}\int \frac{1}{u^2+2u+1} du$$

Step 2: Factor the denominator

Now let's factor the denominator in the integrand: $$u^2+2u+1= (u+1)^2$$ So our integral becomes: $$\frac{1}{2}\int \frac{1}{(u+1)^2} du$$

Step 3: Evaluate the integral

Now our integral can be integrated using the power rule: $$\frac{1}{2}\int \frac{1}{(u+1)^2} du= -\frac{1}{2(u+1)} + C$$ Where C is the constant of integration.

Step 4: Back-substitute the original variable

We made the substitution \(u = x^2\), so let's replace \(u\) with \(x^2\) in our answer: $$-\frac{1}{2(x^2+1)} + C$$ So the integral evaluated is: $$\int \frac{x}{x^{4}+2 x^{2}+1} d x = -\frac{1}{2(x^2+1)} + C$$