Q114 / 180 · KCET 2025 · Mathematics

$\int \frac{\mathrm{dx}}{\mathrm{x}^2\left(\mathrm{x}^4+1\right)^{3 / 4}}$ equals

  • A
    $\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c$
  • B
    $\left(x^4+1\right)^{\frac{1}{4}}+c$
  • C
    $-\left(\mathrm{x}^4+1\right)^{\frac{1}{4}}+\mathrm{c}$
  • Correct answer
    D
    $-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c$

Solution

$\begin{aligned} & \int \frac{\mathrm{dx}}{\mathrm{x}^2\left(\mathrm{x}^4+1\right)^{3 / 4}} \\ & =\int \frac{\mathrm{dx}}{\mathrm{x}^5\left(1+\frac{1}{\mathrm{X}^4}\right)^{3 / 4}} \quad 1+\frac{1}{\mathrm{x}^4}=\mathrm{T} \\ & -4 \mathrm{x}^{-5} \mathrm{dx}=\mathrm{dt} \\ & =-\frac{1}{4} \int \mathrm{t}^{-3 / 4} \mathrm{dt}=-\frac{1}{4} \frac{\mathrm{t}^{1 / 4}}{\frac{1}{4}}+\mathrm{c}=-\left(1+\frac{1}{\mathrm{x}^4}\right)^{1 / 4} \end{aligned}$

Hence, the correct answer is option: 'D'.

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