JEE Mains · Maths · STD 12 - 7.1 indefinite integral
For, \(\alpha, \beta, \gamma, \delta \in N\), if \(\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{ e } x d x=\frac{1}{\alpha}\left(\frac{ x }{ e }\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{ e }{ x }\right)^{\delta x }+ C ,\) Where \(e =\sum \limits_{ n =0}^{\infty} \frac{1}{ n !}\) and \(C\) is constant of integration, then \(\alpha+2 \beta+3 \gamma-4 \delta\) is equal to:
- A \(1\)
- B \(-4\)
- C \(-8\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
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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