Find the integral of y= e^3x / 1+e^x using calculus.

First, you use the substitution u=1+e^x which implies that du=e^x dx. Then, you factorise e^3x = e^2x e^x and replace e^x dx with du. Then by rearranging, e^2x= (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+e^x)²/2 = (1/2)e^2x+e^x+(1/2) , 2u= 2e^x+2 and lnu=ln(1+e^x). So the final answer is ∫y= (1/2)e^2x+ 3e^x+ (5/2) + ln(1+e^x) + c , where c is a constant.