Question

1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)]

Answer 1

∫(1/x^2)cos(1/x)dx

=-∫cos(1/x)d(1/x)

=-sin(1/x)+C

2.

∵(arcsinx)'=xf(x)=(1-x^2)^ (-1/2)

∴f(x)=[x (1-x^2)^ 1/2] ^(-1)

1/f(x)=x(1-x^2) ^1/2

∫1/f(x)dx =∫x(1-x^2) ^1/2dx

=-1/2∫(1-x^2)^ 1/2 d(1-x^2)

= -1/3(1-x^2) ^(3/2) + C

3.∫（lnx/x）dx

=∫（lnx）dlnx

=1/2(lnx)^2+C