Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 21

int(xe^(2x))/(1+2x)^2dx
If f(x) and g(x) are differentiable functions, then
intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx
If we rewrite f(x)=u and g'(x)=v, then
intuvdx=uintvdx-int(u'intvdx)dx
Using the above method of integration by parts,
Let u=xe^(2x)
u'=xd/dx(e^(2x))+e^(2x)d/dx(x)
u'=x(2e^(2x))+e^(2x)
u'=e^(2x)(2x+1)
v=1/(1+2x)^2
intvdx=int(1/(1+2x)^2)dx
Let's integrate by the use of substitution method,
Let t=1+2x
dt=2dx
int(1/(1+2x)^2)dx=intdt/(2t^2)
=1/2(t^(-2+1)/(-2+1))
=-1/(2t)
substitute back t=1+2x,
=-1/(2(1+2x))
int(xe^(2x))/(1+2x)^2dx=xe^(2x)int(1/(1+2x)^2)dx-int(d/dx(xe^(2x))int(1/(1+2x)^2)dx)dx)
=xe^(2x)(-1/(2(1+2x)))-inte^(2x)(1+2x)(-1/(2(1+2x)))dx
=(-xe^(2x))/(2(1+2x))+inte^(2x)/2dx
=(-xe^(2x))/(2(1+2x))+(1/2)e^(2x)/2
=e^(2x)/4-(xe^(2x))/(2(1+2x))
Add a constant C to the solution,
int(xe^(2x))/(1+2x)^2dx=e^(2x)/4-(xe^(2x))/(2(1+2x))+C