Nếu đề là $f(x)=\frac{x-3}{x^2+2x-3}$ thì giải như sau:
Ta có:
\(F(x)=\int f(x)dx=\int \frac{x-3}{x^2+2x-3}dx=\int \frac{\frac{-1}{2}(x+3)+\frac{3}{2}(x-1)}{(x-1)(x+3)}dx\)
\(=-\frac{1}{2}\int \frac{dx}{x-1}+\frac{3}{2}\int \frac{dx}{x+3}=\frac{-1}{2}\ln |x-1|+\frac{3}{2}\ln |x+3|+c\)
\(F(0)=\frac{-1}{2}\ln |0-1|+\frac{3}{2}\ln |0+3|+c=0\Rightarrow c=\frac{-3}{2}\ln 3\)
\(F(2)=\frac{-1}{2}\ln |2-1|+\frac{3}{2}\ln |2+3|-\frac{3}{2}\ln 3=\frac{3}{2}\ln \frac{5}{3}\)
Nhi Le: Ý bạn là \(f(x)=\frac{x-3}{x^2+2x-3}\) hay $f(x)=x-\frac{3}{x^2+2x-3}$ vậy
