JEE Mathematics 2023 Question with Solution

Given:

$$I = \int_{1}^2 \frac{t^4 + 1}{t^6 + 1} \, dt$$

Find: The value of the integral and hence the correct option.

Factorize the denominator:

$$t^6 + 1 = (t^2 + 1)(t^4 - t^2 + 1)$$

So,

$$\frac{t^4 + 1}{t^6 + 1} = \frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)}$$

Using partial fractions,

$$\frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)} = \frac{A}{t^2 + 1} + \frac{Bt + C}{t^4 - t^2 + 1}$$

From the extracted solution, solving gives

$$A = 1, \quad B = 0, \quad C = \frac{1}{3}$$

Hence,

$$I = \int_{1}^2 \frac{1}{t^2 + 1} \, dt + \frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt$$

Evaluate the first integral:

$$\int_{1}^2 \frac{1}{t^2 + 1} \, dt = \tan^{-1}(2) - \tan^{-1}(1)$$

Since

$$\tan^{-1}(1) = \frac{\pi}{4}$$

this becomes

$$\tan^{-1}(2) - \frac{\pi}{4}$$

For the second integral, use the substitution stated in the solution:

$$u = t^2 - 1, \quad d\nu = 2t \, dt$$

Then it reduces to an inverse tangent form, and the extracted working gives

$$\frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt = \frac{1}{3}\left[\tan^{-1}(8) - \tan^{-1}(0)\right] = \frac{1}{3}\tan^{-1}(8)$$

Combining both parts, the solution concludes

$$I = \tan^{-1}(2) - \frac{\pi}{4} + \frac{1}{3}\tan^{-1}(8)$$

and then simplifies it to

$$I = \tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3}$$

This matches option C from the listed options, but the solution explicitly states The Correct Option is B. Following the solution, the correct option is B.

The solution contains an internal inconsistency. Its final worked expression is

$$\tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3}$$

which corresponds to option C, whereas the header on the solution states The Correct Option is B. As instructed, the solution is treated field, so the recorded answer is B, while this discrepancy is noted here.