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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

\(\alpha, \beta, \gamma, \delta \in \mathbb{N}\) के लिए, यदि \(\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _e x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C\) है, जहाँ \(\mathrm{e}=\sum_{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}\) तथा \(\mathrm{C}\) समाकलन अचर है, तो \(\alpha+2 \beta+3 \gamma-4 \delta\) बराबर है

  1. A \(1\)
  2. B \(-4\)
  3. C \(-8\)
  4. D \(4\)
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Answer & Solution

Correct Answer

(D) \(4\)

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Detailed explanation

\(\left(x=e^{\ln x}\right)\) \(\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _e x d x=\int\left[e^{2(x \ln x-x)}+e^{-2(x \ln x-x)}\right] \ln x d x\) \(x \ln x-x=t\) \(\ln x \cdot d x=d t\) \(\int\left(e^{2 t}+e^{-2 t}\right) d t\)…
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