To make the numerator and denominator suitable for substitution, we divide both the numerator and the denominator by x². This is a standard technique for integrals involving x^4+1 or x^4+ax²+1 type denominators.

To make the numerator and denominator suitable for substitution, we divide both the numerator and the denominator by x². This is a standard technique for integrals involving x^4+1 or x^4+ax²+1 type denominators.

We rewrite the denominator in terms of x - (1)/(x) by using the algebraic identity (a-b)² = a² - 2ab + b². In this case, a=x and b=(1)/(x), so a²+b² = (a-b)² + 2ab becomes x² + (1)/(x²) = (x - (1)/(x))² + 2· x · (1)/(x) = (x - (1)/(x))² + 2.

We rewrite the denominator in terms of x - (1)/(x) by using the algebraic identity (a-b)² = a² - 2ab + b². In this case, a=x and b=(1)/(x), so a²+b² = (a-b)² + 2ab becomes x² + (1)/(x²) = (x - (1)/(x))² + 2· x · (1)/(x) = (x - (1)/(x))² + 2.

This integral is of the form ∫ (1)/(a² + x²) dx = (1)/(a) ((x)/(a)). Here, x is u and a² is 2, so a = √2. We apply this formula to evaluate the definite integral with the new limits.

We evaluate the arctan function at the upper and lower limits. As the argument of arctan approaches ∞, the function approaches (π)/(2). As the argument approaches -∞, it approaches -(π)/(2). Subtracting the lower limit from the upper limit, we find the value of the integral.

We evaluate the arctan function at the upper and lower limits. As the argument of arctan approaches ∞, the function approaches (π)/(2). As the argument approaches -∞, it approaches -(π)/(2). Subtracting the lower limit from the upper limit, we find the value of the integral.

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

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

Step-by-step solution

Step 1: Simplify the integrand

To make the numerator and denominator suitable for substitution, we divide both the numerator and the denominator by $x^2$ . This is a standard technique for integrals involving $x^4+1$ or $x^4+ax^2+1$ type denominators.

Step 2: Rewrite the denominator

We rewrite the denominator in terms of $x - \frac{1}{x}$ by using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$ . In this case, $a=x$ and $b=\frac{1}{x}$ , so $a^2+b^2 = (a-b)^2 + 2ab$ becomes $x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2\cdot x \cdot \frac{1}{x} = \left(x - \frac{1}{x}\right)^2 + 2$ .

Step 3: Perform substitution

Let us substitute $u = x - \frac{1}{x}$ . Then, we find the differential $du$ by differentiating $u$ with respect to $x$ . The derivative of $x$ is 1, and the derivative of $-\frac{1}{x}$ (or $-x^{-1}$ ) is $-(-1)x^{-2} = \frac{1}{x^2}$ . This means $du = \left(1 + \frac{1}{x^2}\right) dx$ exactly matches our modified numerator.

Step 4: Change limits of integration

Since we are changing the variable of integration from $x$ to $u$ , we must also change the limits of integration. As $x$ approaches $0$ from the positive side, $u$ approaches $-\infty$ . As $x$ approaches $\infty$ , $u$ also approaches $\infty$ .

Step 5: Substitute into the integral

Substitute $u$ and $du$ into the integral, along with the new limits. The numerator becomes $du$ , and the denominator becomes $u^2 + 2$ . The integral is now a standard form integral.

Step 6: Evaluate the integral

This integral is of the form $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$ . Here, $x$ is $u$ and $a^2$ is $2$ , so $a = \sqrt{2}$ . We apply this formula to evaluate the definite integral with the new limits.

Step 7: Apply the limits of integration

We evaluate the arctan function at the upper and lower limits. As the argument of arctan approaches $\infty$ , the function approaches $\frac{\pi}{2}$ . As the argument approaches $-\infty$ , it approaches $-\frac{\pi}{2}$ . Subtracting the lower limit from the upper limit, we find the value of the integral.

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