MCQ
$\int_{}^{} {\frac{{2x}}{{{{(2x + 1)}^2}}}dx = } $
  • $\frac{1}{2}\log (2x + 1) + \frac{1}{{2(2x + 1)}} + c$
  • B
    $\frac{1}{2}\log (2x + 1) - \frac{1}{{2(2x + 1)}} + c$
  • C
    $2\log (2x + 1) + \frac{1}{{2(2x + 1)}} + c$
  • D
    $2\log (2x + 1) - \frac{1}{{2(2x + 1)}} + c$

Answer

Correct option: A.
$\frac{1}{2}\log (2x + 1) + \frac{1}{{2(2x + 1)}} + c$
a
(a)$\int_{}^{} {\frac{{2x}}{{{{(2x + 1)}^2}}}\,dx = \int_{}^{} {\frac{{2x + 1 - 1}}{{{{(2x + 1)}^2}}}\,dx} } $
$ = \int_{}^{} {\frac{1}{{(2x + 1)}}\,dx} - \int_{}^{} {{{(2x + 1)}^{ - 2}}\,dx} $
$ = \frac{1}{2}\log (2x + 1) + \frac{1}{{2(2x + 1)}} + c$.

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