`int (x e^(2x))/(1 + 2x)^2 dx` Evaluate the integral

`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`

`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`