Answer
$$\int\frac{8x^2+8x+2}{(4x^2+1)^2}dx=\tan^{-1}(2x)-\frac{1}{4x^2+1}+C$$
Work Step by Step
$$A=\int\frac{8x^2+8x+2}{(4x^2+1)^2}dx$$
1) Express the integrand as a sum of partial fractions:
Here we have both an irreducible quadratic factor and a repeated linear factor,
$$\frac{8x^2+8x+2}{(4x^2+1)^2}=\frac{Ax+B}{4x^2+1}+\frac{Cx+D}{(4x^2+1)^2}$$
Clear fractions:
$$(Ax+B)(4x^2+1)+Cx+D=8x^2+8x+2$$ $$4Ax^3+Ax+4Bx^2+B+Cx+D=8x^2+8x+2$$ $$4Ax^3+4Bx^2+x(A+C)+B+D=8x^2+8x+2$$
Equating coefficients of corresponding powers of $y$, we get
$A=0$ and $B=2$
$A+C=8$, so $C=8$
$B+D=2$, so $D=0$
Therefore,
$$\frac{8x^2+8x+2}{(4x^2+1)^2}=\frac{2}{4x^2+1}+\frac{8x}{(4x^2+1)^2}$$
2) Evaluate the integral:
$$A=2\int\frac{dx}{4x^2+1}+\int\frac{8x}{(4x^2+1)^2}dx$$ $$A=\int\frac{d(2x)}{(2x)^2+1}+\int\frac{d(4x^2)}{(4x^2+1)^2}$$ $$A=\tan^{-1}(2x)-\frac{1}{4x^2+1}+C$$
