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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int_{}^{} {\frac{{x{e^x}}}{{{{(1 + x)}^2}}}dx = } $
  • A
    $\frac{{{e^{ - x}}}}{{1 + x}} + c$
  • B
    $ - \frac{{{e^{ - x}}}}{{1 + x}} + c$
  • ✓
    $\frac{{{e^x}}}{{1 + x}} + c$
  • D
    $ - \frac{{{e^x}}}{{1 + x}} + c$
✓

Answer

Correct option: C.
$\frac{{{e^x}}}{{1 + x}} + c$
c
(c)$\int_{}^{} {\frac{{x{e^x}}}{{{{(1 + x)}^2}}}\,dx = \int_{}^{} {\frac{{(x + 1 - 1)}}{{{{(1 + x)}^2}}}{e^x}dx} } $
$ = \int_{}^{} {{e^x}\left( {\frac{1}{{1 + x}} - \frac{1}{{{{(1 + x)}^2}}}} \right)\,dx} = \frac{{{e^x}}}{{1 + x}} + c$.

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