∫[1/（1+x^4）]dx
= 1/2∫[(x^2+1)-(x^2-1)]/（1+x^4）dx
= 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/（1+x^4）dx }
= 1/2 {∫(1+1/x^2)dx /（x^2+1/x^2） - ∫(1-1/x^2)dx/（x^2+1/x^2）}
= 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -2] }
= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - ∫d(x+1/x) /[(x+1/x)^2 -2] }
= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1}
- 1/2√2 ∫d[(x+1/x) /√2] [ 1/{[(x+1/x)/√2] -1} - 1/{[(x+1/x)/√2] +1 }]
= √2/4*arctan[(x-1/x)/√2] - √2/8*ln|(x^2-x√2+1)/(x^2+x√2 +1)| + C
第二个题由分部积分做
因为d((e^x-2)^0.5) = e^x/(2(e^x-2)^0.5) dx
原式=2∫ x d((e^x-2)^0.5)
=2x(e^x-2)^0.5-2∫(e^x-2)^(0.5)dx
令t=(e^x-2)^(1/2) 则dt=e^xdx/2t=(t^2+2)dx/2t
所以dx=2t/(t^2+2) dt
所以∫(e^x-2)^0.5 dx=∫2t^2/(t^2+2) dt
=2∫[1-2/(2+t^2)]dt=2（t-√2arctan(√2/2t))+C
=2(e^x-2)^(1/2)-2√2arctan[√2/2 (e^x-2)^(1/2)]+C
所以原积分=2(x-2)(e^x-2)^(1/2)+4√2arctan[√2/2 (e^x-2)^(1/2)]+C