WBJEE · Maths · Definite Integration
If \(\mathrm{b}=\int_{0}^{1} \frac{\mathrm{e}^{t}}{\mathrm{t}+1} d t\), then \(\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-1}}{\mathrm{t}-\mathrm{a}-1}\) is
- A be \(^{a}\)
- B be \(^{-a}\)
- C \(-\mathrm{be}^{-\mathrm{a}}\)
- D \(-\mathrm{be}^{\mathrm{a}}\)
Answer & Solution
Correct Answer
(C) \(-\mathrm{be}^{-\mathrm{a}}\)
Step-by-step Solution
Detailed explanation
\(\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} d t\) \(=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-(2 \mathrm{a}-1-\mathrm{t})}}{2 \mathrm{a}-1-\mathrm{t}-\mathrm{a}-1} \mathrm{dt}\)…
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