Q90 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
$\int \dfrac{x^4(x^{10} - 1)}{x^{20} + 3x^{10} + 1}\, dx = \cdots$
- A$\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
- Correct answerB$\dfrac{1}{5}\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
- C$\tan^{-1}\left(x^{10} + \dfrac{1}{x^{10}}\right) + c$
- D$\dfrac{1}{10}\tan^{-1}\left(x^{10} + \dfrac{1}{x^{10}}\right) + c$
Solution
$I = \int \dfrac{x^4(x^{10} - 1)}{x^{20} + 3x^{10} + 1}\, dx$
Dividing the numerator and the denominator by $x^{10}$:
$I = \int \dfrac{x^4 - \dfrac{1}{x^6}}{x^{10} + 3 + \dfrac{1}{x^{10}}}\, dx$
Let $x^5 + \dfrac{1}{x^5} = t$
Differentiating with respect to $x$:
$\left(5x^4 - \dfrac{5}{x^6}\right) dx = dt \Rightarrow \left(x^4 - \dfrac{1}{x^6}\right) dx = \dfrac{dt}{5}$
The denominator can be written as:
$x^{10} + \dfrac{1}{x^{10}} + 3 = \left(x^5 + \dfrac{1}{x^5}\right)^2 - 2 + 3 = t^2 + 1$
Substituting these into the integral:
$I = \int \dfrac{1}{t^2 + 1} \dfrac{dt}{5}$
$I = \dfrac{1}{5} \tan^{-1}(t) + c$
$I = \dfrac{1}{5} \tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
Answer: $\dfrac{1}{5}\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
Dividing the numerator and the denominator by $x^{10}$:
$I = \int \dfrac{x^4 - \dfrac{1}{x^6}}{x^{10} + 3 + \dfrac{1}{x^{10}}}\, dx$
Let $x^5 + \dfrac{1}{x^5} = t$
Differentiating with respect to $x$:
$\left(5x^4 - \dfrac{5}{x^6}\right) dx = dt \Rightarrow \left(x^4 - \dfrac{1}{x^6}\right) dx = \dfrac{dt}{5}$
The denominator can be written as:
$x^{10} + \dfrac{1}{x^{10}} + 3 = \left(x^5 + \dfrac{1}{x^5}\right)^2 - 2 + 3 = t^2 + 1$
Substituting these into the integral:
$I = \int \dfrac{1}{t^2 + 1} \dfrac{dt}{5}$
$I = \dfrac{1}{5} \tan^{-1}(t) + c$
$I = \dfrac{1}{5} \tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
Answer: $\dfrac{1}{5}\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
Hence, the correct answer is option: 'B'.