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tegim Teama |
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Answer» ∫ (e^x - 1) / (e^x + 1) dx let u = e^x ; du = e^x dx = ∫ (u - 1) / (u (u + 1)) du = ∫ (2u - (u + 1)) / (u (u + 1)) du = 2 ∫ du / (u + 1) - ∫ du / u = 2 ln(u + 1) - ln(u) + C = 2 ln(e^x + 1) - x + C |
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