MCQ
$\int_{}^{} {\frac{1}{{{x^2}}}{{(2x + 1)}^3}} dx = $
  • $4{x^2} + 12x + 6\log x - \frac{1}{x} + c$
  • B
    $4{x^2} + 12x - 6\log x - \frac{2}{x} + c$
  • C
    $2{x^2} + 8x + 3\log x - \frac{2}{x} + c$
  • D
    $8{x^2} + 6x + 6\log x + \frac{2}{x} + c$

Answer

Correct option: A.
$4{x^2} + 12x + 6\log x - \frac{1}{x} + c$
a
(a)$\int_{}^{} {\frac{1}{{{x^2}}}{{(2x + 1)}^3}dx = \int_{}^{} {\frac{{(8{x^3} + 1 + 12{x^2} + 6x)}}{{{x^2}}}\,dx} } $
$ = \int_{}^{} {\left( {8x + 12 + \frac{6}{x} + \frac{1}{{{x^2}}}} \right)\,dx} = 4{x^2} + 12x + 6\log x - \frac{1}{x} + c$.

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