Question

Evaluate the integral $I=\int \frac{e^{3{\log }_{e}2x}+5e^{2{\log }_{e}2x}}{e^{4{\log }_{e}x}+5e^{3{\log }_{e}x}-7e^{2{\log }_{e}x}}dx,x>0$

• A. ${\log }_e|x^2+5x–7|+c$
• B. $\frac{1}{4}{\log }_e|x^2+5x–7|+c$
• C. $4{\log }_e|x^2+5x–7|+c$
• D. ${\log }_e\sqrt{(x^2+5x–7)}+c$

Answer 1

The correct option is C

$4{\log }_e|x^2+5x–7|+c$

Explanation for the correct option:

Given $I=\int \frac{e^{3{\log }_{e}2x}+5e^{2{\log }_{e}2x}}{e^{4{\log }_{e}x}+5e^{3{\log }_{e}x}-7e^{2{\log }_{e}x}}dx,x>0$

To find the integral of given function,

$I=\int \frac{e^{{\log }_e{\left(2x\right)}^3}+5e^{l{\mathrm{og}}_e{\left(2x\right)}^2}}{e^{{\log }_e{\left(x\right)}^4}+5e^{{\log }_e{\left(x\right)}^3}-7e^{{\log }_e{x}^2}}dx,x>0$

Since, exponential and log function are inverse functions.

$$\begin{gathered} \Rightarrow I=\int \frac{{\left(2x\right)}^3+5{\left(2x\right)}^2}{x^4+5x^3-7x^2}dx \\ =\int \frac{8x^3+20x^2}{x^2\left(x^2+5x-7\right)}dx \\ =\int \frac{4\left(2x+5\right)}{\left(x^2+5x-7\right)}dx \end{gathered}$$

Consider $$\begin{gathered} x^2+5x-7=t \\ \Rightarrow (2x+5)dx=dt \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\int \frac{4}{t}dt \\ \Rightarrow I=4{\log }_e\left|t\right|+c \\ \Rightarrow I=4{\log }_e\left|x^2+5x-7\right|+c \end{gathered}$$

Therefore, the solution of the given integral is $4{\log }_e|x^2+5x–7|+c$

Hence, option (C) is the correct answer.