MCQ
$\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^3}}}{e^x}\;dx = } $
- A$\frac{{ - {e^x}}}{{{{(x + 1)}^2}}} + c$
- ✓$\frac{{{e^x}}}{{{{(x + 1)}^2}}} + c$
- C$\frac{{{e^x}}}{{{{(x + 1)}^3}}} + c$
- D$\frac{{ - {e^x}}}{{{{(x + 1)}^3}}} + c$
✓
Answer
Correct option: B.
$\frac{{{e^x}}}{{{{(x + 1)}^2}}} + c$
b
(b)$\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^3}}}{e^x}dx = \int_{}^{} {{e^x}\left( {\frac{{(x + 1)}}{{{{(x + 1)}^3}}} - \frac{2}{{{{(x + 1)}^3}}}} \right)\,dx} } $
$ = \int_{}^{} {{e^x}\left( {\frac{1}{{{{(x + 1)}^2}}} - \frac{2}{{{{(x + 1)}^3}}}} \right)\,dx} = \frac{{{e^x}}}{{{{(x + 1)}^2}}} + c$.
$\left\{ {\therefore \,\,\,\frac{d}{{dx}}\left( {\frac{1}{{{{(x + 1)}^2}}}} \right) = - \frac{2}{{{{(x + 1)}^3}}}} \right\}$
(b)$\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^3}}}{e^x}dx = \int_{}^{} {{e^x}\left( {\frac{{(x + 1)}}{{{{(x + 1)}^3}}} - \frac{2}{{{{(x + 1)}^3}}}} \right)\,dx} } $
$ = \int_{}^{} {{e^x}\left( {\frac{1}{{{{(x + 1)}^2}}} - \frac{2}{{{{(x + 1)}^3}}}} \right)\,dx} = \frac{{{e^x}}}{{{{(x + 1)}^2}}} + c$.
$\left\{ {\therefore \,\,\,\frac{d}{{dx}}\left( {\frac{1}{{{{(x + 1)}^2}}}} \right) = - \frac{2}{{{{(x + 1)}^3}}}} \right\}$
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