Question

The value of the integral \(\int^2_1\bigg(\frac{t^4+1}{t^6+1}\bigg)\)dt is 

• A. \(tan^{-1}2-\frac{1}{3}tan^{-1}8+\frac{\pi}{3}\)
• B. \(tan^{-1}2-\frac{1}{3}tan^{-1}8-\frac{\pi}{3}\)
• C. \(tan^{-1}\frac{1}{2}+\frac{1}{3}tan^{-1}8-\frac{\pi}{3}\)
• D. \(tan^{-1}\frac{1}{2}-\frac{1}{3}tan^{-1}8+\frac{\pi}{3}\)

Answer 1

The Correct Option is B

(A)The given integral is: \[ \int_{1}^2 \frac{t^4 + 1}{t^6 + 1} \, dt. \] (B)Factorize \( t^6 + 1 \): \[ t^6 + 1 = (t^2 + 1)(t^4 - t^2 + 1). \] Rewrite the integrand: \[ \frac{t^4 + 1}{t^6 + 1} = \frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)}. \] (C)Use partial fractions: \[ \frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)} = \frac{A}{t^2 + 1} + \frac{Bt + C}{t^4 - t^2 + 1}. \] Multiply through and equate terms to solve for \( A, B, C \). Solving gives: \[ A = 1, \, B = 0, \, C = \frac{1}{3}. \] (D)Substituting, the integral becomes: \[ \int_{1}^2 \frac{1}{t^2 + 1} \, dt + \frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt. \] (E)Solve the first term: \[ \int_{1}^2 \frac{1}{t^2 + 1} \, dt = \tan^{-1}(2) - \tan^{-1}(1). \] Since \( \tan^{-1}(1) = \frac{\pi}{4} \), this simplifies to: \[ \tan^{-1}(2) - \frac{\pi}{4}. \] (F) For the second term, perform substitution \( u = t^2 - 1 \), \( du = 2t \, dt \): \[ \frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt = \frac{1}{3} \int \frac{1}{u^2 + 1} \, du = \frac{1}{3} \tan^{-1}(u). \] Back-substitute and evaluate: \[ \frac{1}{3} \left[\tan^{-1}(8) - \tan^{-1}(0)\right] = \frac{1}{3} \tan^{-1}(8). \] (G) Combine the results: \[ \tan^{-1}(2) - \frac{\pi}{4} + \frac{1}{3} \tan^{-1}(8). \] Simplify: \[ \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(8) - \frac{\pi}{3}. \]