To answer your question, use division or fancy algebra (as below).

∫ x^2 dx/(1 + 4x^2)

= (1/4) ∫ 4x^2 dx/(1 + 4x^2)

= (1/4) ∫ [(1 + 4x^2) - 1] dx/(1 + 4x^2)

= (1/4) ∫ [1 - 1/(1 + 4x^2)] dx

= (1/4) ∫ [1 - (1/2) * 2/(1 + (2x)^2)] dx

= (1/4) [x - (1/2) arctan(2x)] + C

= (1/8) [2x - arctan(2x)] + C.

I hope this helps!

You have been careless with brackets, division comes first by default. Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet you have chosen the more complicated way.

Integrate the original integrand by parts:

∫ xtanˉ¹(2x) dx

Let f'(x) = x

f(x) = x² / 2

Let g(x) = tanˉ¹(2x)

g'(x) = 2 / (4x² + 1)

∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx

∫ xtanˉ¹(2x) dx = x²tanˉ¹(2x) / 2 - ∫ x² / (4x² + 1) dx

Integrate the new integrand by algebraic manipulation:

x² / (4x² + 1) = [4x² / (4x² + 1)] / 4

x² / (4x² + 1) = [(4x² + 1 - 1) / (4x² + 1)] / 4

x² / (4x² + 1) = [(4x² + 1) / (4x² + 1) - 1 / (4x² + 1)] / 4

x² / (4x² + 1) = [1 - 1 / (4x² + 1)] / 4

x² / (4x² + 1) = ¼ - 1 / [4(4x² + 1)]

∫ x² / (4x² + 1) dx = ∫ [¼ - 1 / {4(4x² + 1)}] dx

∫ x² / (4x² + 1) dx = ∫ 1 dx / 4 - ∫ 1 / (4x² + 1) dx / 4

∫ x² / (4x² + 1) dx = x / 4 - tanˉ¹(2x) / 8

Put it all together to integrate the original function:

∫ xtanˉ¹(2x) dx = x²tanˉ¹(2x) / 2 - ∫ x² / (4x² + 1) dx

∫ xtanˉ¹(2x) dx = x²tanˉ¹(2x) / 2 - [x / 4 - tanˉ¹(2x) / 8] + C

∫ xtanˉ¹(2x) dx = x²tanˉ¹(2x) / 2 - x / 4 + tanˉ¹(2x) / 8 + C

∫ xtanˉ¹(2x) dx = (4x² + 1)tanˉ¹(2x) / 8 - x / 4 + C