The value of the integral ∫₁² ( (t⁴ + 1)/(t⁶ + 1) )dt is

Solution

Maths sir · 4 steps

Split t⁴+1 as (t⁴-t²+1)+t², and both pieces become tan⁻¹ integrals.

1. Factor the denominator. As a sum of cubes, t⁶ + 1 = (t² + 1)(t⁴ - t² + 1). Write the numerator as t⁴ + 1 = (t⁴ - t² + 1) + t².

   (t⁴+1)/(t⁶+1) = 1/(t²+1) + t²/(t⁶+1)

2. First part. ∫₁² dt/(1+t²) = tan⁻¹ 2 - tan⁻¹ 1 = tan⁻¹ 2 - π/4.
3. Second part. Put u = t³, so du = 3t² dt, and the limits become 1 to 8. This gives 1/3∫₁⁸ du/(1+u²) = 1/3 tan⁻¹ 8 - π/12.
4. So. Adding the two parts gives tan⁻¹ 2 + 1/3 tan⁻¹ 8 - π/4 - π/12 = tan⁻¹ 2 + 1/3 tan⁻¹ 8 - π/3, which is option (c).

The idea to keep. When the denominator is t⁶ + 1, factor it as (t²+1)(t⁴-t²+1). Then split the numerator so that each piece becomes a tan⁻¹ form, either directly or after putting u = t³.

Answer from the NTA final answer key. Solution by PrepWiser.