Question

\(\int\limits^{\sqrt{3}}_1\dfrac{dt}{\left(x^2+1\right)^2}=?\)

Answer 1

Đặt \(x=tant\) suy ra \(dx=\dfrac{dt}{cos^2t}\).
Đổi cận :
\(x=\sqrt{3}\Rightarrow t=arctan\sqrt{3}=\dfrac{\pi}{3}\).
\(x=1\Rightarrow t=arctan1=\dfrac{\pi}{4}\).
\(\int\limits^{\sqrt{3}}_1\dfrac{1}{\left(x^2+1\right)^2}dx\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{dt}{\left(tan^2t+1\right)^2.cot^2t}=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{cos^4t}{cos^2t}dt=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}cos^2tdt=\)\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{1}{2}\left(1+cos2t\right)dt=\)\(=\dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}}{2}-1\right)\).