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• You can rewrite it as e^x(1+e^x)^-1dx Then you can set u = 1 + e^x, giving you: du = e^xdx So you have the integral of (u^-1)du, which is ln(u), which is ln(1 + e^x) + C. The answer is ln(1 + e^x) + C.

  in how do you integrate e^x/1+e^x dx? Comment by gagagoogoololo June 2021
• You can rewrite it as e^x(1+e^x)^-1dx Then you can set u = 1 + e^x, giving you: du = e^xdx So you have the integral of (u^-1)du, which is ln(u), which is ln(1 + e^x) + C. The answer is ln(1 + e^x) + C.

  in how do you integrate e^x/1+e^x dx? Comment by gagagoogoololo June 2021