Question:
The integral $\int \frac{e^{3\log_e{2x}} + 5e^{2\log_e{2x}}}{e^{4\log_e{x}} + 5e^{3\log_e{x}} - 7e^{2\log_e{x}}} \,dx$, x>0, is equal to: (where c is a constant of integration)
The integral $\int \frac{e^{3\log_e{2x}} + 5e^{2\log_e{2x}}}{e^{4\log_e{x}} + 5e^{3\log_e{x}} - 7e^{2\log_e{x}}} \,dx$, x>0, is equal to: (where c is a constant of integration)
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When integrating a rational function, first simplify any complex expressions (like exponential or logarithmic forms). Then, always check if the numerator is a multiple of the derivative of the denominator. If it is, the integral is simply a logarithm.
Updated On: Jan 3, 2026
- $\log_e |x^2+5x-7| + c$
- $4\log_e |x^2+5x-7| + c$
- $\frac{1}{4}\log_e |x^2+5x-7| + c$
- $\log_e \sqrt{x^2+5x-7} + c$
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The Correct Option is B
Solution and Explanation
Simplify numerator:
\[
e^{3\ln(2x)}=(2x)^3=8x^3,\quad
5e^{2\ln(2x)}=20x^2
\]
Denominator:
\[
x^4+5x^3-7x^2
\]
\[
I=\int\frac{8x^3+20x^2}{x^4+5x^3-7x^2}\,dx
=\int\frac{4(2x+5)}{x^2+5x-7}\,dx
\]
Let \(u=x^2+5x-7\Rightarrow du=(2x+5)dx\)
\[
I=4\int\frac{du}{u}
=4\ln|x^2+5x-7|+C
\]
Correct option: (B)
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