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)

{log}_{e} | x^{2} + 5 x - 7 | + c

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\frac{1}{4} {log}_{e} | x^{2} + 5 x - 7 | + c

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4 {log}_{e} | x^{2} + 5 x - 7 | + c

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{log}_{e} \sqrt{x^{2} + 5 x - 7} + c

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Solution

The correct option is C $4 {log}_{e} | x^{2} + 5 x - 7 | + c$
$\int \frac{e^{3 {log}_{e} 2 x + 5 e^{2 {log}_{e} 2 x}}}{e^{4 {log}_{e} x} + 5 e^{3 {log}_{e} x} - 7 e^{2 {log}_{e} x}} d x$

$= \int \frac{8 x^{3} + 5 (4 x^{2})}{x^{4} + 5 x^{3} - 7 x^{2}} d x$

$= \int \frac{8 x + 20}{x^{2} + 5 x - 7} d x$

$= \int \frac{4 (2 x + 5)}{x^{2} + 5 x - 7} d x {\begin{matrix} Let x^{2} + 5 x - 7 = t \Rightarrow (2 x + 5) d x = d t \end{matrix}}$

$= \int \frac{4 d t}{t}$
$= 4 {log}_{e} | t | + C$
$= 4 {log}_{e} | (x^{2} + 5 x - 7) | + c$

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Similar questions

\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,

\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,

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