Question:
The integral $\int \frac{e^{3 \log _{e} 2 x}+5 e^{2 \log _{e} 2 x}}{e^{4 \log _{e} x}+5 e^{3 \log _{e} x}-7 e^{2 \log _{e} x}} d x, x>0$, is equal to: (where $c$ is a constant of integration)
Correct Option: , 3
Solution:
$\int \frac{e^{3 \log _{e} 2 x}+5 e^{2 \log _{e} 2 x}}{e^{4 \log _{e} x}+5 e^{3 \log _{e} x}-7 e^{2 \log _{e} x}} d x$
$=\int \frac{8 x^{3}+5\left(4 x^{2}\right)}{x^{4}+5 x^{3}-7 x^{2}}$
$=\int \frac{8 x^{3}+20 x^{2}}{x^{4}+5 x^{3}-7 x^{2}}$
$=\int \frac{8 x+20}{x^{2}+5 x-7}$
$=\int \frac{4(2 x+5)}{x^{2}+5 x-7}$
$=\int \frac{4 d t}{t}$
$=4 \ln |t|+C$
$=4 \ln \left|\left(x^{2}+5 x-7\right)\right|+c$