Question

一道定积分题，（13）

Answer 1

∫ lnx/(1+x^2)^(3/2) dx
let
x= tanu
dx = (secu)^2 du
∫ lnx/(1+x^2)^(3/2) dx
=∫ lntanu/(secu) du
=∫ lntanu dsinu
=sinu. lntanu -∫ sinu . [(secu)^2/tanu] du
=sinu. lntanu -∫ secu du
=sinu. lntanu -ln|secu+tanu| +C
=[x/√(1+x^2) ]. lnx -ln|√(1+x^2)+x| +C
where
x= tanu
sinu = x/√(1+x^2)

Answer 2

ƒ(x) = { [∫(0→x) sin(t²) dt]/x³，若x ≠ 0
{a，若x = 0
lim(x→0) [∫(0→x) sin(t²) dt]/x³
= lim(x→0) sin(x²)/(3x²)
= (1/3)lim(x²→0) sin(x²)/x²
= 1/3 • 1
= 1/3

∵ƒ(x)在x = 0处连续
∴a = 1/3