Q114 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

$\int \frac{1}{x^m \sqrt[m]{x^m+1}} d x=$

  • A
    $\frac{1}{m-1}\left(\frac{\sqrt[m]{x^m+1}}{x}\right)^m+c$
  • Correct answer
    B
    $\frac{-1}{m-1}\left(\frac{\sqrt[m]{x^m+1}}{x}\right)^{m-1}+c$
  • C
    $\frac{-1}{m}\left(\frac{\sqrt[m]{x^m+1}}{x}\right)^m+c$
  • D
    $\frac{1}{m}\left(\frac{\sqrt[m-1]{x^m+1}}{x}\right)^m+c$

Solution

$\mathrm{I}=\int \frac{1}{x^m\left(x^m+\right)^{\frac{1}{m}}} d x$ Let $u=\mathrm{x}^{1-\mathrm{m}} ; d u=\frac{1-m}{x^m} d x$ $x^m=u^{\frac{m}{1-m}}$ $\mathrm{I}=-\frac{1}{m-1} \int \frac{d u}{\left(u^{\frac{m}{1-m}})+1\right)^{\frac{1}{m}}}=\frac{-1}{m-1} \int \frac{u^{\frac{1}{m-1}}}{\left(u^{\frac{m}{m-1}}+1\right)^{\frac{1}{m}}} d u$
Let $u^{\frac{m}{m-1}}+1=v \Rightarrow d v=mu^\frac{{\frac{m}{m-1}}-1}{m-1}-d u$
$\mathbb{I}=\frac{-1}{m} \int \frac{1}{v^{\frac{1}{m}}} d v \Rightarrow \mathbb{I}=\frac{-v^{1-\frac{1}{m}}}{(1-\frac{1}{m})m}$ $\begin{aligned} & \mathrm{I}=\frac{\left(u^{\frac{m}{m-1}}+1\right)^{1-\frac{1}{m}}}{\left(1-\frac{1}{m}\right)^m}+\mathrm{C} ; \mathrm{I}=\frac{-\left(x^{\frac{(1-m) m}{m-1}}+1\right)^{\mathfrak{1}-\frac{1}{m}}}{\left(1-\frac{1}{m}\right) m}+\mathrm{C} \\ & \mathrm{I}=\frac{-1}{m-1}\left(\frac{\sqrt[m]{x^m+1}}{x}\right)^{m-1}+\mathrm{C}\end{aligned}$

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

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