Q90 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
$\int\dfrac{x^4 + 1}{x^6 + 1}dx = $
- A$\tan^{-1}x - \dfrac{1}{3}\tan^{-1}(x^3) + c$
- Correct answerB$\tan^{-1}x + \dfrac{1}{3}\tan^{-1}(x^3) + c$
- C$\tan^{-1}x - \tan^{-1}(x^3) + c$
- D$\tan^{-1}x + \tan^{-1}(x^3) + c$
Solution
The given integral is $I = \int \dfrac{x^4 + 1}{x^6 + 1} dx$
The denominator can be factored as $x^6 + 1 = (x^2)^3 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$
Rewriting the numerator as $x^4 + 1 = (x^4 - x^2 + 1) + x^2$
$I = \int \dfrac{(x^4 - x^2 + 1) + x^2}{(x^2 + 1)(x^4 - x^2 + 1)} dx$
Separating the terms:
$I = \int \dfrac{x^4 - x^2 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} dx + \int \dfrac{x^2}{x^6 + 1} dx$
$I = \int \dfrac{1}{x^2 + 1} dx + \dfrac{1}{3} \int \dfrac{3x^2}{(x^3)^2 + 1} dx$
Substituting $t = x^3 \Rightarrow dt = 3x^2 dx$ in the second integral:
$I = \tan^{-1}x + \dfrac{1}{3} \int \dfrac{1}{t^2 + 1} dt$
$I = \tan^{-1}x + \dfrac{1}{3} \tan^{-1}(t) + c$
$I = \tan^{-1}x + \dfrac{1}{3} \tan^{-1}(x^3) + c$
Answer: $\tan^{-1}x + \dfrac{1}{3}\tan^{-1}(x^3) + c$
The denominator can be factored as $x^6 + 1 = (x^2)^3 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$
Rewriting the numerator as $x^4 + 1 = (x^4 - x^2 + 1) + x^2$
$I = \int \dfrac{(x^4 - x^2 + 1) + x^2}{(x^2 + 1)(x^4 - x^2 + 1)} dx$
Separating the terms:
$I = \int \dfrac{x^4 - x^2 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} dx + \int \dfrac{x^2}{x^6 + 1} dx$
$I = \int \dfrac{1}{x^2 + 1} dx + \dfrac{1}{3} \int \dfrac{3x^2}{(x^3)^2 + 1} dx$
Substituting $t = x^3 \Rightarrow dt = 3x^2 dx$ in the second integral:
$I = \tan^{-1}x + \dfrac{1}{3} \int \dfrac{1}{t^2 + 1} dt$
$I = \tan^{-1}x + \dfrac{1}{3} \tan^{-1}(t) + c$
$I = \tan^{-1}x + \dfrac{1}{3} \tan^{-1}(x^3) + c$
Answer: $\tan^{-1}x + \dfrac{1}{3}\tan^{-1}(x^3) + c$
Hence, the correct answer is option: 'B'.