MCQ
$\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}\,dx = } $
- A$\frac{e}{4}$
- ✓$\frac{e}{4} - 1$
- C$\frac{e}{4} + 1$
- DNone of these
✓
Answer
Correct option: B.
$\frac{e}{4} - 1$
b
(b) $\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}} dx = \int_0^1 {\frac{{{e^x}(x + 1 - 2)}}{{{{(x + 1)}^3}}}\,} dx$
(b) $\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}} dx = \int_0^1 {\frac{{{e^x}(x + 1 - 2)}}{{{{(x + 1)}^3}}}\,} dx$
$\int_0^1 {\frac{{{e^x}}}{{{{(x + 1)}^2}}}} dx - 2\int_0^1 {\frac{{{e^x}}}{{{{(x + 1)}^3}}}} dx = \left[ {\frac{{{e^x}}}{{{{(x + 1)}^2}}}} \right]_0^1 $
$= \frac{e}{4} - 1$.
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