Question

Integral (x e^2x)/(2x+1)^2

Answer 1

Integration by Parts.
It's not immediately obvious but this numerator differentiates very nicely.

$\rm \left(x e^{2x}\right)'&=e^{2x}+2x e^{2x}=e^{2x}(2x+1)$

Notice that taking the derivative of the numerator actually creates a factor of our denominator.

Ah ha! So we've found a good choice for our u,

$\rm u=x e^{2x}\qquad\qquad\qquad dv=\frac{1}{(2x+1)^2}~dx\\ \\ du=e^{2x}(2x+1)dx\qquad v=\frac{-1}{2(2x+1)}$

letting dv be everything else.

Applying Integration by parts

$\rm =(u)(v)-\int (v)du$

gives us,

$\rm =\left(x e^{2x}\right)\left(\dfrac{-1}{2(2x+1)}\right)-\int \left(\dfrac{-1}{2(2x+1)}\right)e^{2x}}(2x+1)dx$

Simplifying things a little bit before integrating again,

$\rm =-\dfrac{x e^{2x}}{2(2x+1)}+\frac12\int e^{2x}~dx$

and integrating the last term,

$\rm =-\dfrac{x e^{2x}}{2(2x+1)}+\frac14 e^{2x}$

looking for a common denominator, multiplying the first term by 2/2 and the second term by (2x+1)/(2x+1),

$\rm =\dfrac{-2x e^{2x}}{4(2x+1)}+\dfrac{e^{2x}(2x+1)}{4(2x+1)}$

Combine the fractions together, factor out the exponential,

$\rm =\dfrac{e^{2x}(-2x+2x+1)}{4(2x+1)}$

combine like-terms as a final step,
and include a constant of integration,

$\rm =\dfrac{e^{2x}}{4(2x+1)}+c$