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Problem If I = ∫ e x e 4 x + e 2 x + 1 d x I = \int \frac{e^x}{e^{4x} + e^{2x} + 1} dx I = ∫ e 4 x + e 2 x + 1 e x d x J = ∫ e − x e − 4 x + e − 2 x + 1 d x J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx J = ∫ e − 4 x + e − 2 x + 1 e − x d x c c c J − I J - I J − I 1 2 log ∣ e 4 x − e 2 x + 1 e 4 x + e 2 x + 1 ∣ + C \frac{1}{2} \log \left| \frac{e^{4x} - e^{2x} + 1}{e^{4x} + e^{2x} + 1} \right| + C 2 1 log e 4 x + e 2 x + 1 e 4 x − e 2 x + 1 + C 1 2 log ∣ e 2 x + e x + 1 e 2 x − e x + 1 ∣ + C \frac{1}{2} \log \left| \frac{e^{2x} + e^x + 1}{e^{2x} - e^x + 1} \right| + C 2 1 log e 2 x − e x + 1 e 2 x + e x + 1 + C 1 2 log ∣ e 2 x − e x + 1 e 2 x + e x + 1 ∣ + C \frac{1}{2} \log \left| \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right| + C 2 1 log e 2 x + e x + 1 e 2 x − e x + 1 + C 1 2 log ∣ e 4 x + e 2 x + 1 e 4 x − e 2 x + 1 ∣ + C \frac{1}{2} \log \left| \frac{e^{4x} + e^{2x} + 1}{e^{4x} - e^{2x} + 1} \right| + C 2 1 log e 4 x − e 2 x + 1 e 4 x + e 2 x + 1 + C Browse the problem library