KCET · Maths · Indefinite Integration
The value of \( \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} \mathrm{dx} \) is equal to
- A \( 00 \)
- B \( \frac{x^{3}}{3} \)
- C \( \frac{3}{x^{3}} \)
- D \( \frac{1}{x} \)
Answer & Solution
Correct Answer
(B) \( \frac{x^{3}}{3} \)
Step-by-step Solution
Detailed explanation
Given that, \( \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x \)
We know that \( e^{k \log x}=x^{k} \)
\[
\begin{array}{l}
\text { So, } I=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \\
=\int \frac{x^{5}(x-1)}{x^{3}(x-1)} d x=\int x^{2} d x \\
=\frac{x^{3}}{3}+c
\end{array}
\]
We know that \( e^{k \log x}=x^{k} \)
\[
\begin{array}{l}
\text { So, } I=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \\
=\int \frac{x^{5}(x-1)}{x^{3}(x-1)} d x=\int x^{2} d x \\
=\frac{x^{3}}{3}+c
\end{array}
\]
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