Question

Evaluate the indefinite integral: \( \int \frac{x^2+1}{x^4+1}\,dx \)

• A. \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C \)
• B. \( \tan^{-1}x+C \)
• C. \( \frac{1}{2}\ln(x^2+1)+C \)
• D. \( \frac{x}{x^2+1}+C \)

Hint:

Whenever you see \( x^4 + 1 \) in the denominator paired with an \( x^2 \) term in the numerator, always divide by \( x^2 \) immediately. It's a hallmark NCERT advanced integration type.

Answer 1

The Correct Option is A

Concept: Integrals of the form \( \int \frac{x^2 \pm 1}{x^4 + kx^2 + 1}\,dx \) are evaluated by dividing both the numerator and the denominator by \( x^2 \), followed by an algebraic substitution of the form \( u = x \mp \frac{1}{x} \).

Step 1: Divide both the numerator and denominator by \( x^2 \). Dividing each term inside the integral yields: \[ I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}}\,dx \]

Step 2: Rewrite the denominator and apply integration by substitution. We can rewrite the denominator expression as a perfect square: \[ x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 \] Now, let \( u = x - \frac{1}{x} \). Differentiating both sides with respect to \( x \): \[ du = \left(1 + \frac{1}{x^2}\right)dx \] Substituting these values back into the integral: \[ I = \int \frac{du}{u^2 + (\sqrt{2})^2} \]

Step 3: Apply the standard integral formula and substitute back. Using the standard form \( \int \frac{1}{u^2+a^2}\,du = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) \): \[ I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C \] Substituting \( u = \frac{x^2-1}{x} \) back into the expression: \[ I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) + C \]