Integrals: PYQ Walkthrough (14)

Question

Evaluate

$\int \frac{x^2 + 1}{x^4 + 1}\,dx$

This appeared in JEE Main 2023 and tests the “divide by $x^2$” trick.

Solution — Step by Step

Final answer: $\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^2 - 1}{x\sqrt{2}}\right) + C$.

Why This Works

The trick is to spot that the numerator is the derivative of $(x - 1/x)$ and the denominator can be expressed as a quadratic in $(x - 1/x)$. This converts an unfamiliar integrand into the standard form $1/(u^2 + a^2)$.

The same trick works for similar integrals of the form $(x^2 \pm 1)/(x^4 + kx^2 + 1)$ — divide by $x^2$ and look for $x \pm 1/x$ substitutions.

Alternative Method

Partial fractions over $\mathbb{R}$: factor $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$, decompose, and integrate each piece. This works but takes 5x longer — JEE doesn’t reward it.

Common Mistake

Not recognising that $x^4 + 1 = (x^2)^2 + 1$ doesn’t factor nicely as $(x^2+1)(x^2 - 1)$ — that gives $x^4 - 1$. Always check by expansion before going down the wrong path.