Question

Tìm nguyên hàm các hàm số hữu tỉ sau :

a.) \(\int\frac{1}{x^2-3x+2}dx\)

 

b) \(\int\frac{1}{4x^2-3x-1}dx\)

Answer 1

a) \(\int\frac{1}{x^2-3x+2}dx=\frac{1}{2-1}\int\frac{1}{\left(x-1\right)\left(x-2\right)}dx\)

=\(\int\frac{1}{x-2}dx-\int\frac{1}{x-1}dx=ln\left|x-2\right|-ln\left|x-1\right|=ln\left|\frac{x-2}{x-1}+C\right|\)

 

b) \(\int\frac{1}{4x^2-3x-1}dx=\frac{1}{4}.\frac{1}{\left(1-\frac{1}{4}\right)}\int\frac{1}{\left(x+\frac{1}{4}\right)\left(x-1\right)}dx\)

=\(\frac{1}{3}.\left[\int\frac{1}{x-1}dx-\int\frac{1}{x+\frac{1}{4}}dx\right]\)

=\(\frac{1}{3}\left[ln\left|x-1\right|-ln\left|x+\frac{1}{4}\right|\right]=\frac{1}{3}ln\left|\frac{x-1}{x+\frac{1}{4}}\right|+C\)

=\(\frac{1}{3}ln\left|\frac{4\left(x-1\right)}{4x+1}+C\right|\)