Find the integral of y= e^3x / 1+e^x using calculus.
First, you use the substitution u=1+ex which implies that du=ex dx. Then, you factorise e3x = e2x ex and replace ex dx with du. Then by rearranging, e2x= (u-1)2 , so you are now ready to substitute the x s with u s. Therefore, the integration becomes ∫(u-1)2 /u du. Then by expanding and simplifying the brackets the integral becomes ∫u-2+ (1/u) du so we now integrate and the integral becomes u2/2 -2u + lnu + C'. But then, we need to substitute the x's back into the equation so u2/2 = (1+ex)2/2 = (1/2)e2x+ex+(1/2) , 2u= 2ex+2 and lnu=ln(1+ex). So the final answer is ∫y= (1/2)e2x+ 3ex+ (5/2) + ln(1+ex) + c , where c is a constant.
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