97松農-積分
此題我花六小時還是不得其解[img]http://farm4.static.flickr.com/3321/3565993289_21b3f0dcea_b.jpg[/img]
請老師撥冗解題
不勝感荷 \[\int \frac{x + 1}{\sqrt {4 - 6x - x^2} }dx = \int \frac{(x + 3) - 2}{\sqrt {13 - (x+3)^2} }dx \]
（令 \(t=x+3\Rightarrow dt = dx\)）
\[ = \int \frac{t-2}{\sqrt {13-t^2}} dt\]
\[ = \int \frac{t}{\sqrt {13-t^2}} dt - \int \frac{2}{\sqrt {13-t^2}} dt\]
\[ = \int -\frac{d(13-t^2)}{2\sqrt {13-t^2}} - 2 \int \frac{1}{\sqrt{13}\cdot\sqrt {1-\left(\frac{t}{\sqrt{13}}\right)^2}} dt\]
\[ = -\sqrt{13-t^2} - 2 \int \frac{1}{\sqrt {1-\left(\frac{t}{\sqrt{13}}\right)^2}} d\left(\frac{t}{\sqrt{13}}\right)\]
（令 \(\displaystyle\frac{t}{\sqrt{13}}=\sin\theta \Rightarrow d\left(\frac{t}{\sqrt{13}}\right) = \cos\theta\, d\theta\)）
\[ = -\sqrt{13-t^2} - 2 \int \frac{1}{\sqrt {1-\sin^2\theta}}\cos\theta d\theta\]
\[ = -\sqrt{13-t^2} - 2 \int d\theta\]
\[ = -\sqrt{13-t^2} - 2 \theta + c\]
\[ = -\sqrt{4-6x-x^2} - 2 \sin^{-1}\left(\frac{t}{\sqrt{13}}\right) + c\]
\[ = -\sqrt{4-6x-x^2} - 2 \sin^{-1}\left(\frac{x+3}{\sqrt{13}}\right) + c\]
其中， \(c\) 為常數。 老師的積分功力超強
佩服 謝謝您
教安
頁:
[1]