ta co:
[tex] x^4+1= (x^2+x sqrt{2}+1)(x^2-x sqrt{2}+1) [/tex]
gia su:
[tex]\frac{x^2-1}{x^4+1} =[/tex] [tex]\frac{Ax+B}{x^2-x sqrt{2}+1}+\frac{Cx+D}{x^2+x sqrt{2}+1}[/tex]
Đồng nhất 2 vế, khi đó:
[tex] A=\frac{1}{sqrt{2}}, B=-\frac{1}{2}, C=-\frac{1}{sqrt{2}}, D=-\frac{1}{2} [/tex]
thay vào :
[tex]\int\limits_{}^{}\frac{x^2-1}{x^4+1}dx =[/tex] [tex]\int\limits_{}^{}\frac{\frac{1}{sqrt{2}}x-\frac{1}{2}}{x^2-x sqrt{2}+1}dx+\int\limits_{}^{}\frac{-\frac{1}{sqrt{2}}x-\frac{1}{2}}{x^2-x sqrt{2}+1}dx[/tex]
[tex]<=> \int\limits_{}^{}\frac{x^2-1}{x^4+1}dx=[/tex] [tex]\frac{1}{2 sqrt{2}}(\int\limits_{}^{} \frac{\2x-x sqrt{2}}{x^2-x sqrt{2}+1}dx-\int\limits_{}^{}\frac{2x+x sqrt{2}}{x^2+x sqrt{2}+1}dx )[/tex]
[tex]<=> \int\limits_{}^{}\frac{x^2-1}{x^4+1}dx=[/tex] [tex]\frac{1}{2 sqrt{2}}(\int\limits_{}^{} \frac{d(x^2-x sqrt{2}+1)}{x^2-x sqrt{2}+1}-\int\limits_{}^{}\frac{d(x^2+x sqrt{2}+1)}{x^2+x sqrt{2}+1} )[/tex]
[tex]<=> \int\limits_{}^{}\frac{x^2-1}{x^4+1}dx=[/tex] [tex]\frac{1}{2 sqrt{2}} ( ln(x^2-x sqrt{2}+1)-ln(x^2+x sqrt{2}+1) ) [/tex]
[tex]<=> \int\limits_{}^{}\frac{x^2-1}{x^4+1}dx=[/tex] [tex]\frac{1}{2 sqrt{2}}ln\mid\frac{x^2-x sqrt{2}+1}{x^2+x sqrt{2}+1}\mid + C[/tex]
sặc! bài này khó quá...
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
