\( \int\displaystyle\frac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx,\ x\gt0 \) and \( x\ne1 \) is equal to (consider \( \log x=\log_{e}x \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{x^{3}}{3}+C \), where \( C \) is an arbitrary constant.
Since \( e^{n\log x}=x^{n} \), the exponentials become powers of x.
\( \displaystyle \frac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}=\frac{x^{7}-x^{6}}{x^{5}-x^{4}} \)
\( \displaystyle =\frac{x^{6}(x-1)}{x^{4}(x-1)} \)
Cancelling \( x-1 \), which is allowed since \( x\ne1 \),
\( \displaystyle =x^{2} \)
Hence
\( \displaystyle \int x^{2}\,dx=\frac{x^{3}}{3}+C \)