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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}\,dx = } $
  • A
    $\frac{e}{4}$
  • ✓
    $\frac{e}{4} - 1$
  • C
    $\frac{e}{4} + 1$
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{e}{4} - 1$
b
(b) $\int_0^1 {\frac{{{e^x}(x - 1)}}{{{{(x + 1)}^3}}}} dx = \int_0^1 {\frac{{{e^x}(x + 1 - 2)}}{{{{(x + 1)}^3}}}\,} dx$

$\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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