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