MCQ
$\int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}\,dx}$ is equal to
  • A
    $\frac{{{e^x}}}{{\left( {x - 1} \right)}} + c$
  • $\frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
  • C
    $\frac{{{-e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
  • D
    $\frac{{{2e^x}}}{{{{\left( {x - 1} \right)}^3}}} + c$

Answer

Correct option: B.
$\frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
b
$\int e^{x}\left\{\frac{(x-1)-2}{(x-1)^{3}}\right\} d x$

$=\int e^{x}\left\{\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\right\} d x$

$=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}-1)^{2}}+\mathrm{c}$

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