怎么积分,详细过程
3个回答
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1/[(1-t^2)t^4]
=(1-t^4+t^4)/[(1-t^2)t^4]
=(1-t^4)/[(1-t^2)t^4]+t^4/[(1-t^2)t^4]
=(1-t^2)(1+t^2)/[(1-t^2)t^4]+1/(1-t^2)
=(1+t^2)/t^4+1/(1-t^2)
=1/t^4+1/t^2+1/[(1-t)(1+t)]
=1/t^4+1/t^2+1/[2(1-t)]+1/[2(1+t)]
1/t^4的积分为-1/(3t^3)+C;
1/t^2的积分为-1/(2t)+C;
1/[2(1-t)]的积分为-(1/2)ln(t-1)+C;
1/[2(1+t)]的积分为(1/2)ln(t+1)+C;
so,1/[(1-t^2)t^4]的积分为-1/(3t^3)-1/(2t)+(1/2)ln[(t+1)/(t-1)]+C
=(1-t^4+t^4)/[(1-t^2)t^4]
=(1-t^4)/[(1-t^2)t^4]+t^4/[(1-t^2)t^4]
=(1-t^2)(1+t^2)/[(1-t^2)t^4]+1/(1-t^2)
=(1+t^2)/t^4+1/(1-t^2)
=1/t^4+1/t^2+1/[(1-t)(1+t)]
=1/t^4+1/t^2+1/[2(1-t)]+1/[2(1+t)]
1/t^4的积分为-1/(3t^3)+C;
1/t^2的积分为-1/(2t)+C;
1/[2(1-t)]的积分为-(1/2)ln(t-1)+C;
1/[2(1+t)]的积分为(1/2)ln(t+1)+C;
so,1/[(1-t^2)t^4]的积分为-1/(3t^3)-1/(2t)+(1/2)ln[(t+1)/(t-1)]+C
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