We can rewrite the denominator x² + (1)/(x²) using the algebraic identity (a-b)² = a² - 2ab + b². Here, a=x and b=(1)/(x), so (x - (1)/(x))² = x² - 2(x)((1)/(x)) + (1)/(x²) = x² - 2 + (1)/(x²). Therefore, x² + (1)/(x²) = (x - (1)/(x))² + 2.

We can rewrite the denominator x² + (1)/(x²) using the algebraic identity (a-b)² = a² - 2ab + b². Here, a=x and b=(1)/(x), so (x - (1)/(x))² = x² - 2(x)((1)/(x)) + (1)/(x²) = x² - 2 + (1)/(x²). Therefore, x² + (1)/(x²) = (x - (1)/(x))² + 2.

The integral transforms into a standard form ∫ (1)/(u² + a²) du = (1)/(a) tan^(-1)((u)/(a)) + C. Here, a² = 2, so a = √2. We then evaluate the definite integral using the new limits.

We evaluate the inverse tangent function at the limits. As u → ∞, tan^(-1)(u) → (π)/(2). As u → -∞, tan^(-1)(u) → -(π)/(2). Substituting these values gives the final result.

Evaluate the definite integral of ( $x^2 + 1$ ) / ( $x^4 + 1$ ) dx from 0 to infinity.

Answer: $\frac{\pi}{\sqrt{2}}$

Step-by-step solution

Step 1: Rewrite the integrand

To simplify the integrand, we divide both the numerator and the denominator by $x^2$ . This transformation is useful for integrals involving $x^4 + 1$ in the denominator, as it often leads to a form that can be integrated using substitution.

Step 2: Manipulate the denominator

We can rewrite the denominator $x^2 + \frac{1}{x^2}$ using the algebraic identity (a-b)^2 = $a^2 - 2ab + b^2$ . Here, $a=x$ and $b=\frac{1}{x}$ , so (x - $\frac{1}{x}$ )^2 = $x^2 - 2(x)(\frac{1}{x}) + \frac{1}{x^2} = x^2 - 2 + \frac{1}{x^2}$ $. Therefore, x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$ .

Step 3: Apply substitution

We introduce a substitution $u = x - \frac{1}{x}$ . Differentiating u with respect to x gives du = (1 + $\frac{1}{x^2}$ ) dx, which matches the numerator. We also need to change the limits of integration. As x approaches 0 from the positive side, u approaches - $\infty$ $. As x approaches \infty$ , u approaches $\infty$ .

Step 4: Integrate with respect to u

The integral transforms into a standard form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$ $. Here, a^2 = 2$ , so $a = \sqrt{2}$ . We then evaluate the definite integral using the new limits.

Step 5: Evaluate the definite integral

We evaluate the inverse tangent function at the limits. As u $\to \infty$ $, \tan^{-1}(u) \to \frac{\pi}{2}$ . As u $\to -\infty$ $, \tan^{-1}(u) \to -\frac{\pi}{2}$ . Substituting these values gives the final result.

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Evaluate the definite integral of (x2+1x^2 + 1x2+1) / (x4+1x^4 + 1x4+1) dx from 0 to infinity.