 | (1)
10/{(x+1)(x^2+4)} = a/(x+1)+(bx+c)/(x^2+4)とおきます。
10 = a(x^2+4)+(bx+c)(x+1) = (a+b)x^2+(b+c)x+(4a+c)
a+b = 0・・・(1)
b+c = 0・・・(2)
4a+c = 10・・・(3)
(1)よりb = -a, (1)(2)よりc = -b = a
(3)より4a+a = 10 ⇒ a = 2, b = -2, c = 2
∫10dx/{(x+1)(x^2+4)} = ∫{2/(x+1)+(-2x+2)/(x^2+4)}dx
= 2log(x+1)-log(x^2+4)+arctan(x/2)+C
# ∫{-2x/(x^2+4)}dx = -∫{(x^2+4)'/(x^2+4)}dx = -log(x^2+4)
# ∫{2/(x^2+4)}dxは、x = 2*tan(t)とおいて置換積分します。
(2)
x^2/{(x^2+1)(x^2+4)} = (ax+b)/(x^2+1)+(cx+d)/(x^2+4)とおきます。
x^2 = (ax+b)(x^2+4)+(cx+d)(x^2+1) = (a+c)x^3+(b+d)x^2+(4a+c)x+(4b+d)
a+c = 0・・・(1)
b+d = 1・・・(2)
4a+c = 0・・・(3)
4b+d = 0・・・(4)
(1)よりc = -a, (3)より4a-a = 0 ⇒ a = 0, c = 0
(2)よりd = 1-b, (4)より4b+(1-b) = 0 ⇒ b = -1/3, d = 1-(-1/3) = 4/3
∫x^2dx/{(x^2+1)(x^2+4)} = ∫{(-1/3)/(x^2+1)+(4/3)/(x^2+4)}dx
= -1/3*arctan(x)+2/3*arctan(x/2)+C
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