Q113 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

$\int \frac{1}{x^2}(2 x+1)^3 d x$ is equal to

  • Correct answer
    A
    $4 x^2+12 x+6 \log x-\frac{1}{x}+C$
  • B
    $4 x^2+12 x-6 \log x-\frac{2}{x}+C$
  • C
    $2 x^2+8 x+3 \log x-\frac{2}{x}+C$
  • D
    $8 x^2+6 x+6 \log x+\frac{2}{x}+C$

Solution

We have, $\int \frac{1}{x^2}(2 x+1)^3 d x$ $\begin{aligned} & =\int \frac{\left(8 x^3+1+12 x^2+6 x\right)}{x^2} d x \\ & =\int\left(8 x+12+\frac{6}{x}+\frac{1}{z^2}\right) d x\end{aligned}$ $\left[\begin{array}{c}\because \int x^n d x=\frac{x^{n+1}}{n+1}+c, n \neq-1 \\ \int \frac{d x}{x}=\log x+c, x>0\end{array}\right]$ $=4 x^2+12 x+6 \log x-\frac{1}{x}+C$

Hence, the correct answer is option: 'A'.

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