1+x^4 = (1+x²)² - 2x² = (1+x²+√2x)(1+x²-√2x)1/(1+x^4) = [1/(1+x²-√2x) - 1/(1+x²+√2x)]/2√2x = 1/2√2 *[1/x + (√2-x)/(1+x²-√2x) - 1/x + (√2+x)/(1+x²+√2x)] = 1/4√2 * [(2x+2√2)/(x²+√2x+1) - (2x-2√2)/(x²+1-√2x)] = 1/4√2 *[(2x+√2)/(x²+√2x+1) - (2x-√2)/(x²+1-√2x) + √2/(x²+√2x+1) + √2/(x²+1-√2x)]对(2x+√2)/(x²+√2x+1)求积分得ln(x²+√2x+1)对(2x-√2)/(x²+1-√2x)求积分得ln(x²+1-√2x)对√2/(x²+√2x+1)求积分得2arctan(√2x+1)对√2/(x²-√2x+1)求积分得2arctan(√2x-1)原式 = 1/4√2 *{ln[(x²+√2x+1))/(x²+1-√2x)] + 2arctan(√2x+1) + 2arctan(√2x-1)} + C
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回答时间：2024-04-24 00:33
