Question

求不定積分∫

dx

(1+ex )2

．

Answer 1

令1+e^x=t，
則dt=e^xdx=（t-1）dx，dx=

dt

t-1

.
∴∫

dx

(1+ex)2

=∫

dt

(t-1)t2

=∫(

1

t(t-1)

-

1

t2

)dt
=∫(

1

t-1

-

1

t

-

1

t2

)dt
=ln(t-1)-lnt+

1

t

+C
=lnex-ln(1+ex)+

1

1+ex

+C
=x-ln(1+ex)+

1

1+ex

+C．