MHT CET · Maths · Indefinite Integration
\(\int \frac{\mathrm{d} x}{2 \mathrm{e}^{2 x}+3 \mathrm{e}^x+1}=\)
- A \(x+\log \left(\mathrm{e}^x+1\right)-2 \log \left(2 \mathrm{e}^x+1\right)+\mathrm{c},\) where \(c\) is the constant of integration
- B \(x-\log \left(\mathrm{e}^x+1\right)+4 \log \left(\mathrm{e}^x+1\right)+\mathrm{c}\) where \(c\) is the constant of integration
- C \(x+\log \left(\mathrm{e}^x+1\right)-4 \log \left(2 \mathrm{e}^x+1\right)+\mathrm{c},\) where \(c\) is the constant of integration
- D \(x-\log \left(\mathrm{e}^x+1\right)+2 \log \left(2 \mathrm{e}^x+1\right)+\mathrm{c},\) where \(c\) is the constant of integration
Answer & Solution
Correct Answer
(A) \(x+\log \left(\mathrm{e}^x+1\right)-2 \log \left(2 \mathrm{e}^x+1\right)+\mathrm{c},\) where \(c\) is the constant of integration
Step-by-step Solution
Detailed explanation
\(u = \mathrm{e}^x \implies \mathrm{d}x = \frac{\mathrm{d}u}{u}\) \(\int \frac{\mathrm{d}u}{u(2u^2+3u+1)} = \int \frac{\mathrm{d}u}{u(u+1)(2u+1)}\)
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