JEE Mains · Maths · STD 12 - 7.1 indefinite integral
Let \(I(x)=\int \frac{(x+1)}{x\left(1+x e^x\right)^2} d x, x > 0\), If \(\lim _{x \rightarrow \infty} I(x)=0\), then \(I(1)\) is equal to
- A \(\frac{e+1}{e+2}-\log _e(e+1)\)
- B \(\frac{e+1}{e+2}+\log _e(e+1)\)
- C \(\frac{e+2}{e+1}+\log _e(e+1)\)
- D \(\frac{e+2}{e+1}-\log _e(e+1)\)
Answer & Solution
Correct Answer
(D) \(\frac{e+2}{e+1}-\log _e(e+1)\)
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