Answer
$$\int\frac{16x^3}{4x^2-4x+1}dx=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$
Work Step by Step
$$I=\int\frac{16x^3}{4x^2-4x+1}dx$$
1) Perform long division:
We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{16x^3}{4x^2-4x+1}=4x+4+\frac{12x-4}{4x^2-4x+1}=4x+4+\frac{12x-4}{(2x-1)^2}$$
2) Express the remainder fraction as a sum of partial fractions:
$$\frac{12x-4}{(2x-1)^2}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}$$
Clear fractions:
$$A(2x-1)+B=12x-4$$ $$2Ax-A+B=12x-4$$
Equating coefficients of corresponding powers of $x$, we get
$2A=12$, so $A=6$
$-A+B=-4$, so $B=-4+A=2$
Therefore,
$$\frac{12x-4}{(2x-1)^2}=\frac{6}{2x-1}+\frac{2}{(2x-1)^2}$$
2) Evaluate the integral:
$$I=\int(4x+4)dx+\int\frac{6}{2x-1}dx+\int\frac{2}{(2x-1)^2}dx$$ $$I=2x^2+4x+\int\frac{3}{2x-1}d(2x-1)+\int\frac{1}{(2x-1)^2}d(2x-1)$$ $$I=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$
