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provide solution for RD Sharma maths class 12 chapter indefinite integrals exercise 18.1 question 3

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Answer: \frac{x^{3}}{3}+c

Hint: Use the formulas of logarithm

Given:\int \frac{e^{6\log_{e}x}-e^{5\log_{e}x}}{e^{4\log_{e}x}-e^{3\log_{e}x}}dx

Solution:\int \frac{e^{6\log_{e}x}-e^{5\log_{e}x}}{e^{4\log_{e}x}-e^{3\log_{e}x}}dx

=\int \int \frac{e^{\log_{e}x^{6}}-e^{\log_{e}x^{5}}}{e^{\log_{e}x^{4}}-e^{\log_{e}x^{3}}}dx                                        \left ( \because x\log_{e}x=\log_{e}x^{x} \right )

=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}}dx=\int \frac{x^{5}(x-1)}{x^{3}(x-1)}dx                            \left ( \because e^{\log e^{x}}=x \right )

=\int \frac{x^{5}}{x^{3}}dx=\int x^{2}dx

=\frac{x^{3}}{3}+c

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