Question

∫(3+2t)/2(1-t^2)dt

Answer 1

问：∫(3+2t)/2(1-t^2)dt

答：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