Question

求定积分∫（0-1） 1/1+x^4dx

Answer 1

先算不定积分∫（x²+1）/(x^4+1)dx
分子分母同时除以x²得：
原式=∫（1+1/x²）/（x²+1/x²）dx
=∫d（x-1/x)/[（x-1/x）²+2]
=1/√2 arctan[(x-1/x)/√2]+C
=1/√2 arctan[(x²-1)/√2x]+C
同理可解得：
不定积分∫（x²-1）/(x^4+1)dx
=1/2√2 ln|(x²-√2x+1)/(x²+√2x+1)|+C

∴不定积分∫1/(x^4+1)dx
=1/2∫[（x²+1）-(x²-1)]/(x^4+1)dx
=1/2∫（x²+1）/(x^4+1)dx-1/2∫（x²-1）/(x^4+1)dx
=1/2√2 arctan[(x²-1)/√2x]-1/4√2 ln|(x²-√2x+1)/(x²+√2x+1)|+C
再代入上下限0和1得：
原定积分=[π-ln（3-2√2）]/4√2
可能有算错的地方哈

Answer 2

∫(0→1) dx/(1 + x⁴)
= (1/2)∫(0→1) [(x² + 1) - (x² - 1)]/(1 + x⁴) dx
= (1/2)∫(0→1) (x² + 1)/(x⁴ + 1) dx - (1/2)∫(0→1) (x² - 1)/(x⁴ + 1) dx
= (1/2)∫(0→1) (1 + 1/x²)/(x² + 1/x²) dx - (1/2)∫(0→1) (1 - 1/x²)/(x² + 1/x²) dx
= (1/2)∫(0→1) d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫(0→1) d(x + 1/x)/[(x + 1/x)² - 2]
= (1/2) · 1/√2 · arctan[(x - 1/x)/√2] - (1/2) · 1/(2√2) · ln|[(x + 1/x) - √2]/[(x + 1/x) + √2]|：(0→1)
= 1/(2√2) · arctan[x/√2 - 1/(√2x)] - 1/(4√2) · ln|(x² - √2x + 1)/(x² + √2x + 1)|：(0→1)
= 1/(2√2) · [0 - (- π/2)] - 1/(4√2) · [ln(3 - 2√2) - 0]
= [π - ln(3 - 2√2)]/(4√2)