どうしても解けないので教えてください。
(1) x + 2
∫ ――――― dx
4
x - 1
(2) 3 2
x-3x-4x-20
∫ ―――――――――― dx
4
x -16
(3)∫xcosxsinxdx
どうしても解けないので教えてください。
(1) x + 2
∫ ――――― dx
4
x - 1
(2) 3 2
x-3x-4x-20
∫ ―――――――――― dx
4
x -16
(3)∫xcosxsinxdx
(1) (x + 2)/(\(x^{4}\) - 1) = a/(x - 1) + b/(x + 1) + (cx + d)/(\(x^{2}\) + 1)
と置く。 両辺を \(x^{4}\) - 1 倍すると
x + 2 = a(x + 1)(\(x^{2}\) + 1) + b(x - 1)(\(x^{2}\) + 1)
+ (cx + d)(x + 1)(x - 1) …[a]
これに x = 1 を代入
3 = 4a ∴a = \(\frac{3}{4}\).
x = -1 を代入
1 = -4b ∴b = -\(\frac{1}{4}\)
式 [a] から
4(cx + d)(x + 1)(x - 1)
= 4(x + 2) - 3(x + 1)(\(x^{2}\) + 1) + (x - 1)(\(x^{2}\) + 1)
= 4(x + 2) - 2(\(x^{2}\) + 1)(x + 2)
= -2(x + 2)(\(x^{2}\) + 1 - 2)
= -2(x + 2)(\(x^{2}\) - 1).
従って
2(cx + d) = -(x + 2)
即ち c = -\(\frac{1}{2}\), d = -1.
従って
与式 = (\(\frac{3}{4}\))∫dx/(x - 1) -(\(\frac{1}{4}\))∫dx/(x + 1)
- (\(\frac{1}{4}\))∫(2xdx)/(\(x^{2}\) + 1) -∫dx/(\(x^{2}\) + 1)
= (\(\frac{3}{4}\))log|x - 1| - (\(\frac{1}{4}\))log|x + 1| - (\(\frac{1}{4}\))log|\(x^{2}\) + 1|
- Tan^(-1) x + C, C: 積分定数。
(2) (\(x^{3}\) - 3\(x^{2}\) - 4x - 20)/(\(x^{4}\) - 16)
= a/(x - 2) + b/(x + 2) + (cx + d)/(\(x^{2}\) + 4)
と置く。両辺を \(x^{4}\) - 16 倍すると
\(x^{3}\) - 3\(x^{2}\) - 4x - 20
= a(x + 2)(\(x^{2}\) + 4) + b(x - 2)(\(x^{2}\) + 4) + (cx + d)(x - 2)(x + 2) …[b]
これに x = 2 を代入。
-32 = 32a ∴a = -1.
x = -2 を代入。
-32 = -32b ∴b = 1.
式 [b] から
(cx + d)(x - 2)(x + 2)
= \(x^{3}\) - 3\(x^{2}\) - 4x - 20 + (x + 2)(\(x^{2}\) + 4) - (x - 2)(\(x^{2}\) + 4)
= \(x^{3}\) - 3\(x^{2}\) - 4x - 20 + 4(\(x^{2}\) + 4)
= \(x^{3}\) + \(x^{2}\) - 4x - 4
= (x + 1)\(x^{2}\) - 4(x + 1)
= (x + 1)(\(x^{2}\) - 4).
故に
cx + d = x + 1
即ち c = d = 1.
よって
与式 = - ∫dx/(x - 2) + ∫dx/(x + 2) + (\(\frac{1}{2}\))∫(2xdx)/(\(x^{2}\) + 4)
+ ∫dx/(\(x^{2}\) + 4)
= - log|x - 2| + log|x + 2| + (\(\frac{1}{2}\))log|\(x^{2}\) + 4| + (\(\frac{1}{2}\))Tan^(-1)(\(\frac{x}{2}\)) + C,
C: 積分定数。
ここで最後の積分は x = 2tanθ と置くと
∫dx/(\(x^{2}\) + 4) =∫(2dθ/co\(s^{2}\)θ)/(4ta\(n^{2}\)θ + 4)
= (\(\frac{1}{2}\))∫dθ/((ta\(n^{2}\)θ + 1)co\(s^{2}\)θ)
= (\(\frac{1}{2}\))∫dθ = (\(\frac{1}{2}\))θ
だから。
(3) 与式 = (\(\frac{1}{2}\))∫x sin 2x dx
= (\(\frac{1}{2}\))∫x d((-\(\frac{1}{2}\))cos 2x)
= (\(\frac{1}{2}\))[-(\(\frac{x}{2}\))cos 2x + (\(\frac{1}{2}\))∫cos 2x dx]
= -(\(\frac{x}{4}\))cos 2x + (\(\frac{1}{8}\))sin 2x + C. C: 積分定数。