MCQ
$\int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}\,\,dx = \,\,} $
  • A
    $\frac{1}{{{{(x + 4)}^2}}} + c$
  • B
    $\frac{{{e^x}}}{{{{(x + 4)}^2}}} + c$
  • $\frac{{{e^x}}}{{x + 4}} + c$
  • D
    $\frac{{{e^x}}}{{x + 3}} + c$

Answer

Correct option: C.
$\frac{{{e^x}}}{{x + 4}} + c$
c
(c) $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx} $$I = \int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}dx} $
$ \Rightarrow I = \int {{e^x}\,\left( {\frac{1}{{x + 4}} - \frac{1}{{{{(x + 4)}^2}}}} \right)\,dx} $
$\therefore I = \frac{{{e^x}}}{{x + 4}} + c$.

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