let 1+ 4x^2 = t
then dt/dx= 8x dx
x / (1 + 4x^2)^2 dx = 1/8 (1/t^2) dt
integral = - 1/8(1/t) dt +C = - 1/(8(1+4x^2))+ C as integral of 1/t^2 or t^-2 is -1/t
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let 1+ 4x^2 = t
then dt/dx= 8x dx
x / (1 + 4x^2)^2 dx = 1/8 (1/t^2) dt
integral = - 1/8(1/t) dt +C = - 1/(8(1+4x^2))+ C as integral of 1/t^2 or t^-2 is -1/t