적분 하나

Q.

다음 정적분을 구하라.

(*) $ \int_{0}^{1} \frac{x^2 + 3x + 1}{x^4 + x^2 + 1}~dx $

A.

(*) 

\begin{aligned} \int_{0}^{1}{ \frac{x^2 + 3x + 1}{x^4 + x^2 + 1}~dx} = & \int_{0}^{1} \frac{1+\frac{3}{x} + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} dx \\ =& \int_{0}^{1} {\frac{1+\frac{1}{x^2}}{ ( \frac{1}{x} - x )^2 +3} dx} + \int_0^1 \frac{3x}{x^4+ x^2 + 1} dx \\ =& \int_{0}^{1} {\frac{1+\frac{1}{x^2}}{ ( \frac{1}{x} - x )^2 +3} dx} + \frac{3}{2} \int_0^1 \frac{2x}{x^4+ x^2 + 1} dx  \end{aligned}

$ Let~~ u= \frac{1}{x} - x,~~ v= x^2 $ then, $ du= -(1+ \frac{1}{x^2})dx,~~dv = 2x~dx $ 

\begin{aligned} (\textrm{준식}) =& - \int_{\infty}^0 (u^2 + 3)^{-1}~du + \frac{3}{2} \int_0^1 (v^2 + v + 1)^{-1}~dv \\ =& \int_{0}^{\infty} (u^2 + 3)^{-1}~du + \frac{3}{2} \int_0^1 (v^2 + v + 1)^{-1}~dv \\ =& \frac{1}{3} \lim_{t \to \infty} \int_0^t \frac{1}{(\frac{u}{\sqrt{3}})^2 +1}~du + \frac{3}{2} \cdot \frac{4}{3} \int_0^1 \frac{1}{(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}})^2 +1}~dv \end{aligned}

앞 항과 뒷 항을 나눠서 하자. 각각 식을 $ f(u),~g(v) $라고 명명한다. 먼저 앞 항부터 $ Let~~ \tan \theta = \frac{u}{\sqrt{3}}, ~~then~~ \theta = \arctan(\frac{u}{\sqrt{3}}),~~ du = \sqrt{3} \sec^2 \theta ~d \theta $

$ f(u) = \frac{\sqrt{3}}{3} \lim_{t \to \infty} \int_0^t \frac{\sec^2 \theta}{\tan^2 +1}d \theta $이고, 분자와 분모가 소거된다.  $ f(u) = \frac{\sqrt{3}}{3} \lim_{t \to \infty} \int_0^t d \theta = \frac{\sqrt{3}}{3} \lim_{t \to \infty} \theta ~|_0^t   $가 된다. 여기서 $ \theta $는 다시 $ \arctan (\frac{u}{\sqrt{3}}) $로 치환되고 이를 정적분하면 된다. $ \arctan(\frac{x}{\sqrt{3}}) $이 $ \frac{\pi}{2} $인 것을 이용하자.

$$ \therefore ~f(u)  = \frac{\sqrt{3}}{3} \lim_{t \to \infty} \arctan(\frac{x}{\sqrt{3}}) ~|_0^t = \frac{\pi}{2\sqrt{3}}  $$

이번엔 $ g(v) $를 해보자.

$ Let~~ \tan \theta = \frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}, ~~then~~ v = - \frac{1}{2} + \frac{\sqrt{3}}{2} \tan \theta,~~ dv = \frac{\sqrt{3}}{2} \sec^2 \theta ~d \theta $

$ g(v) = \frac{3}{2} \cdot \frac{2\sqrt{3}}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 \theta}{\tan^2 +1}d \theta $이고, 분자와 분모가 소거된다.  $ g(v) = \frac{3}{2} \cdot \frac{2 \sqrt{3}}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d \theta = \frac{3}{2} \cdot \frac{2 \sqrt{3}}{3}  \theta ~|_{\frac{\pi}{6}}^{\frac{\pi}{3}} $가 된다.

$$ \therefore ~g(v)  = \frac{3}{2} \cdot \frac{2 \sqrt{3}}{3} \cdot \frac{\pi}{6} = \frac{\pi}{2 \sqrt{3}}  $$

$$ \therefore~ (\textrm{준식}) = f(u) + g(v) = \frac{\pi}{\sqrt{3}} $$