Question

Evaluate: \[ \int \dfrac{x^2+1}{x^4+1}\,dx \]

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

Hint:

Whenever expressions like \(x^4+1\) appear in integrals, first factorize into quadratic factors before attempting substitution or partial fractions.

Answer 1

The Correct Option is A

Concept: Integrals involving fourth-degree polynomials are usually simplified by factorization into irreducible quadratic factors. After simplification, standard trigonometric inverse forms are identified.

Step 1: Observing the denominator carefully.
Given: \[ I= \int \dfrac{x^2+1}{x^4+1}\,dx \] Notice that: \[ x^4+1 \] cannot be factorized directly over real linear factors. Hence we factorize into quadratic factors.

Step 2: Factorizing \(x^4+1\).
Using the standard identity: \[ x^4+1 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1) \] Thus, \[ I= \int \dfrac{x^2+1} { (x^2+\sqrt2x+1)(x^2-\sqrt2x+1) } dx \]

Step 3: Using partial fraction decomposition.
Assume: \[ \dfrac{x^2+1} { (x^2+\sqrt2x+1)(x^2-\sqrt2x+1) } = \dfrac{Ax+B}{x^2+\sqrt2x+1} + \dfrac{Cx+D}{x^2-\sqrt2x+1} \] Multiplying throughout: \[ x^2+1 = (Ax+B)(x^2-\sqrt2x+1) + (Cx+D)(x^2+\sqrt2x+1) \] After comparing coefficients and simplifying, we obtain: \[ A=-\frac{1}{2\sqrt2}, \qquad B=\frac12, \qquad C=\frac{1}{2\sqrt2}, \qquad D=\frac12 \] Thus, \[ I= \int \left[ \dfrac{ -\frac{x}{2\sqrt2}+\frac12 } {x^2+\sqrt2x+1} + \dfrac{ \frac{x}{2\sqrt2}+\frac12 } {x^2-\sqrt2x+1} \right] dx \]

Step 4: Converting into standard inverse tangent form.
After algebraic simplification and combining suitable terms, the integral reduces to: \[ I= \dfrac{1}{\sqrt2} \tan^{-1} \left( \dfrac{x^2-1}{\sqrt2x} \right) +C \]

Step 5: Verification by differentiation.
Differentiate: \[ \dfrac{1}{\sqrt2} \tan^{-1} \left( \dfrac{x^2-1}{\sqrt2x} \right) \] Using: \[ \dfrac{d}{dx}(\tan^{-1}u) = \dfrac{u'}{1+u^2} \] and simplifying carefully gives: \[ \dfrac{x^2+1}{x^4+1} \] Hence the integration is verified. Therefore, \[ \boxed{ \int \dfrac{x^2+1}{x^4+1}\,dx = \dfrac{1}{\sqrt2} \tan^{-1} \left( \dfrac{x^2-1}{\sqrt2x} \right) +C } \]