Q114 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

$\int\left[\frac{x^4-x}{x^{20}}\right]^{1 / 4} d x=$

  • Correct answer
    A
    $\frac{4}{15}\left(\frac{\left(x^3-1\right)^5}{x^{15}}\right)^{1 / 4}+C$
  • B
    $\frac{4}{15}\left(\frac{x^4+1}{x^4}\right)^{1 / 4}+C$
  • C
    $\frac{\sqrt{x^4+x^2+1}}{x}+C$
  • D
    $\frac{3}{4}\left(x^{4 / 3}+x^{1 / 3}\right)+C$

Solution

Let $\begin{aligned} I & =\int\left(\frac{x^4-x}{x^{20}}\right)^{1 / 4} d x \\ I & =\int \frac{1}{x^4}\left(1-\frac{1}{x^3}\right)^{1 / 4} d x\end{aligned}$ $\begin{aligned} & \text { put } \quad 1-\frac{1}{x^3}=t \Rightarrow \frac{3}{x^4} d x=d t \\ & \because \quad I=\frac{1}{3} \int(t)^{1 / 4} d t \Rightarrow I=\frac{4}{15} \cdot t^{5 / 4}+C \\ & I=\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5 / 4}+C \Rightarrow I=\frac{4}{15}\left(\frac{\left(x^3-1\right)^5}{x^{15}}\right)^{1 / 4}+C\end{aligned}$

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

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