Mathematics

$$\displaystyle \int { \dfrac { { e }^{ 5\log { x }  }-{ e }^{ 4\log { x }  } }{ { e }^{ 3\log { x }  }-{ e }^{ 2\log { x }  } }  } dx=\_ \_ \_ \_ \_ \_ +c$$


ANSWER

$$\dfrac{{ x }^{ 3 }} 3$$


SOLUTION
Given,

$$\displaystyle \int \dfrac{e^{5\ln \left(x\right)}-e^{4\ln \left(x\right)}}{e^{3\ln \left(x\right)}-e^{2\ln \left(x\right)}}dx$$

Now,

$$\displaystyle \dfrac{e^{5\ln \left(x\right)}-e^{4\ln \left(x\right)}}{e^{3\ln \left(x\right)}-e^{2\ln \left(x\right)}}$$

$$\displaystyle=\dfrac{e^{4\ln \left(x\right)}\left(e^{\ln \left(x\right)}-1\right)}{e^{3\ln \left(x\right)}-e^{2\ln \left(x\right)}}$$

$$\displaystyle=\dfrac{e^{4\ln \left(x\right)}\left(e^{\ln \left(x\right)}-1\right)}{e^{2\ln \left(x\right)}\left(e^{\ln \left(x\right)}-1\right)}$$

$$\displaystyle=\dfrac{e^{4\ln \left(x\right)}}{e^{2\ln \left(x\right)}}$$

$$\displaystyle=\dfrac{e^{4\ln \left(x\right)}}{x^2}$$

$$\displaystyle=\dfrac{x^4}{x^2}=x^2$$

$$\displaystyle \therefore \int \dfrac{e^{5\ln \left(x\right)}-e^{4\ln \left(x\right)}}{e^{3\ln \left(x\right)}-e^{2\ln \left(x\right)}}dx=\int x^2 dx$$

$$=\dfrac{x^{2+1}}{2+1}$$

$$=\dfrac{x^3}{3}+C$$
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Single Correct Medium Published on 17th 09, 2020
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