∫(0->1)1/(x^2-2)^2dx
1个回答

∫(0→1) 1/(x² - 2)² dx

= ∫(0→1) 1/[(x + √2)²(x - √2)²] dx

= ∫(0→1) { 1/[8√2(x + √2)] + 1/[8(x + √2)²] + 1/[8√2(√2 - x)] + 1/[8(x - √2)²] } dx

= [1/(8√2)]ln[(x + √2)/(x - √2)] - 1/[8(x + √2)] - 1/[8(√2 - x)] |(0→1)

= (√2/16)ln[(1 + √2)/(√2 - 1)] - 1/[8(1 + √2)] - 1/[8(1 - √2)] + 1/(8√2) - 1/(8√2)

= (√2/16)ln(3 + 2√2) + 1/4

≈ 0.4058