Question

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

• A. $\log _{ e }\left| x ^{2}+5 x -7\right|+ c$
• B. $4 \log _{e}\left|x^{2}+5 x-7\right|+c$
• C. $\frac{1}{4} \log _{ e }\left| x ^{2}+5 x -7\right|+ c$
• D. $\log _{e} \sqrt{x^{2}+5 x-7}+c$

Answer 1

Answer: (B)

$\int \frac{ e ^{3 \log _{ e } 2 x }+5 e ^{2 \log _{ e } 2 x }}{ e ^{4 \log _{ e } x }+5 e ^{3 \log _{ e } x }-7 e ^{2 \log _{ e } x }} dx , x >0$
$=\int \frac{(2 x )^{3}+5(2 x )^{2}}{ x ^{4}+5 x ^{3}-7 x ^{2}} d x =\int \frac{4 x ^{2}(2 x +5)}{ x ^{2}\left( x ^{2}+5 x -7\right)} dx$
$=4 \int \frac{ d \left( x ^{2}+5 x -7\right)}{\left( x ^{2}+5 x -7\right)}=4 \log _{ e }\left| x ^{2}+5 x -7\right|+ c$