For $x > 1, \int\frac{e^{7\log x}-e^{5\log x}}{e^{5\log x}-e^{4\log x}} dx$ equals.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{x^3}{3}+\frac{x^2}{2}+ C$: C is a constant of integration
Given: $\int \frac{e^{7\log x} - e^{5\log x}}{e^{5\log x} - e^{4\log x}}\,dx$, for $x > 1$
Use the identity: $e^{a\log x} = x^a$
$\Rightarrow \int \frac{x^7 - x^5}{x^5 - x^4}\,dx$
Factor numerator and denominator:
$= \int \frac{x^5(x^2 - 1)}{x^4(x - 1)}\,dx$
$= \int \frac{x^5(x - 1)(x + 1)}{x^4(x - 1)}\,dx$
$= \int \frac{x^5(x + 1)}{x^4}\,dx$
Simplify:
$= \int x(x + 1)\,dx = \int (x^2 + x)\,dx$
Integrate:
$= \int x^2\,dx + \int x\,dx = \frac{x^3}{3} + \frac{x^2}{2} + C$