Question

∫ ( e^5 logx - e^3logx ÷e^4logx-e^2logx ) dx = ?

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Answer 1

People also ask

What is the 9th square number?

81

"Is it possible to recreate this up to 12x12?"

0 Squared = 0

6 Squared = 36

7 Squared = 49

8 Squared = 64

9 Squared = 81

8 more rows

Answer 2

Answer:

ex + C or x²/2 + C

Step-by-step-explanation:

$\int \frac{e {}^{5}logx - {e}^{3} logx}{ {e}^{4}logx - {e}^{2}logx } dx \\ \\ \int \frac{e {}^{3}logx( {e}^{2} - 1)}{ {e}^{2} logx( {e}^{2} - 1) } dx \\ \\ \int e \: dx \\ \\ ex + C$

If your question is: $\int \frac{e^{5 logx} - e^{3logx}}{e^{4logx} - e^{2logx}} dx$

$\implies\int \frac{e^{5 log_e x} - e^{3log_ex}}{e^{4log_ex} - e^{2log_ex}} dx \\ \\ \implies\int \frac{x^{5 log_e e} - x {}^ {3log_ee}}{x^{4log_ee} - x^{2log_ee}} dx \\ \\ \implies \int \frac{ {x}^{5(1) } - {x}^{3(1)} }{ {x}^{4(1)} - {x}^{2(1)} } dx$

$\implies \int \frac{ { {x}^{} }^{5} - {x}^{3} }{ {x}^{4} - {x}^{2} } dx \\ \\ \implies \int \frac{ {x} ( {x}^{4} - {x}^{2} )}{ x {}^{4} - {x}^{2}} dx$

$\\ \implies\int x \: dx \\ \\\implies \frac{ {x}^{2} }{2} + C$