Question

求∫dt/2t(1+t^2)上1下4

Answer 1

定积分的上、下限难以表示,下面用[t=1→4]来表示.
∫dt/2t(1+t^2) [t=1→4]
=∫[1/2t(1+t^2)]dt [t=1→4]
=∫[1/t[(t^2)/2+1/2] [t=1→4]
查积分表,有：
∫dx/[x(ax^2+b)=(1/2b)ln(x^2/|ax^2+b|)+C
对比题目,可知：x=t、a=1/2、b=1/2
所以：
∫[1/t[(t^2)/2+1/2] [t=1→4]
=ln[t^2/|(1/2)t^2+1/2|]+C [t=1→4]
=ln(16/|8+1/2|)-ln(1/|1/2+1/2|)
=ln(32/17)-ln1
=ln(32/17)
=5ln2-ln17