Question

\(\int\dfrac{4x^2+3x+1}{1-2x}dx\) bằng

Answer 1

Lời giải:
\(\int \frac{4x^2+3x+1}{1-2x}dx=-\int \frac{4x^2+3x+1}{2x-1}dx=-\frac{1}{2}\int \frac{8x^2+6x+2}{2x-1}dx\)

\(=-\frac{1}{2}\int \frac{4x(2x-1)+5(2x-1)+7}{2x-1}dx\)

\(=-\frac{1}{2}\int (4x+5+\frac{7}{2x-1})dx\)

\(=-\frac{1}{2}\int (4x+5)dx-\frac{1}{2}\int \frac{7}{2x-1}dx\)

\(=-\frac{1}{2}\int (4x+5)dx-\frac{7}{4}\int \frac{d(2x-1)}{2x-1}\)

\(=-\frac{1}{2}(2x^2+5x)-\frac{7}{4}\ln |2x-1|+c\)