Question

请问（xe^x)/(x+1)^2的不定积分怎么求?

Answer 1

为表述方便,记不定积分符号为J

则J（xe^x)/(x+1)^2dx=J （xe^x)/(x+1)^2d(x+1)

= -J (xe^x)d(1/(x+1))

分步积分=-(xe^x)/(x+1)+J 1/(x+1)d(xe^x)

== -(xe^x)/(x+1)+J 1/(x+1)e^x(x+1)dx

=-(xe^x)/(x+1)+J e^xdx

=-(xe^x)/(x+1)+e^x+C

C为常数