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

x^{2} \neq n π + 1, n \in N

\int x \sqrt{\frac{2 sin (x^{2} - 1) - sin 2 (x^{2} - 1)}{2 sin (x^{2} - 1) + sin 2 (x^{2} - 1)}} d x

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, x > 0,

(\frac{- 8 p}{7})

∣ ∣ x^{2} - 5 x + 7 - p ∣ ∣ = 6 + ∣ ∣ x^{2} - 5 x + 1 - p ∣ ∣

x \in [- 1, 3]

∣ ∣ x^{2} - 5 x + 7 ∣ ∣ + ∣ ∣ x^{2} - 5 x - 14 ∣ ∣ = 21

x^{2} + 5 x - 7 = 0

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