[toÁn 12] nguyÊn hÀm

H

hmu95

$ b, I=\int \frac{1}{x^2+2x+5}=\int \frac{1}{(x+1)^2+4}$
Đặt
$x+1=2tant$ \Rightarrow $dx=2(tan^2t+1)dt$

[tex]\Rightarrow I=\frac{1}{2}\int dt=\frac{t}{2}+C=\frac{arctan\frac{x+1}{2}}{2}+C[/tex]
 
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