The integral $\int \frac{\mathrm{e}^{3 \log _{e} 2 x}+5 \mathrm{e}^{2 \log _{e} 2 x}}{\mathrm{e}^{4 \log _{e} x}+5 \mathrm{e}^{3 \log _{e} x}-7 \mathrm{e}^{2 \log _{e} x}} \mathrm{dx}, x>0$,
is equal to :
(where $c$ is a constant of integration)
Correct Option: , 2
$\int \frac{\mathrm{e}^{3 \log _{\mathrm{e}} 2 \mathrm{x}}+5 \mathrm{e}^{2 \log _{\mathrm{e}} 2 \mathrm{x}}}{\mathrm{e}^{4 \log _{\mathrm{e}} \mathrm{x}}+5 \mathrm{e}^{3 \log _{\mathrm{e}} \mathrm{x}}-7 \mathrm{e}^{2 \log _{\mathrm{e}} \mathrm{x}}} \mathrm{dx}, \mathrm{x}>0$
$=\int \frac{(2 x)^{3}+5(2 x)^{2}}{x^{4}+5 x^{3}-7 x^{2}} d x=\int \frac{4 x^{2}(2 x+5)}{x^{2}\left(x^{2}+5 x-7\right)} d x$
$=4 \int \frac{d\left(x^{2}+5 x-7\right)}{\left(x^{2}+5 x-7\right)}=4 \log _{e}\left|x^{2}+5 x-7\right|+c$
option (2)
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