Let u = e^{6x} + 9. Then du = 6e^{6x} dx and e^{6x} dx = 1 6 du, so e^{6x} e^{6x} + 9 dx = 1 u 1 6 du = 1 6 ln|u| + C = 1 6 ln(e^{6x} + 9) + C. Is this correct? edit: my syntax is messed up sorry I will fix
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∫▒〖e^6x/(e^6x+9) dx〗 let e^(6x )=u Therefore 6e^6x dx=du hence ∫▒〖e^6x/(e^6x+9) dx〗 = 1/6 ∫▒〖1/(u+9) du〗 Now you can finish it
Let Now plugging in and we get: You can check the answer here: int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha
Originally Posted by x3bnm Let Now plugging in and we get: You can check the answer here: int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha And of course, since for all x, so is , so the absolute value signs here are overkill
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