|
|
a) $I=\int\limits_{0}^{1}\frac{1}{2x^{2}+5x+2}dx=\frac{1}{2}\int\limits_{0}^{1}\frac{1}{(x+\frac{5}{4})^{2}-\frac{9}{16}}dx$
Dat: $x+\frac{5}{4}=t\Rightarrow dx=dt$. Doi can: $x=0\Rightarrow t=\frac{5}{4}$ ; $x=1\Rightarrow t=\frac{9}{4}$.
Ta thu duoc tich phan moi:
$\frac{1}{2}\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t^{2}-\frac{9}{16}}dt=\frac{1}{3}\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t-\frac{3}{4}}dt-\int\limits_{\frac{5}{4}}^{\frac{9}{4}}\frac{1}{t+\frac{3}{4}}dt$. Den day la tich phan co ban.
|