09/06/10 02:31:47
>>30
∫{6c^2/(x^3+c^3)}dx
=2[∫{1/(x+c)}dx-∫{(x-2c)/(x^2-cx+c^2)}dx]
=2log(x+c)-∫{(2x-c)/(x^2-cx+c^2)}dx-∫{(-3c)/(x^2-cx+c^2)}dx
=2log|x+c|-log|x^2-cx+c^2|+3c∫{1/((x-c/2)^2+(3/4)c^2)}dx
=2log|x+c|-log|x^2-cx+c^2|+3c∫{1/((x-c/2)^2+((√3)/2)c)^2}dx
=2log|x+c|-log|x^2-cx+c^2|+3c(2/((√3)c))arctan((2/((√3)c))(x-c/2))
=2log|x+c|-log|x^2-cx+c^2|+2(√3)arctan((2x-c)/((√3)c))
∫{1/(x^3+c^3)}dx
＝{2log|x+c|-log|x^2-cx+c^2|+2(√3)arctan((2x-c)/((√3)c))}/(6c^2)
この式でc^3=b/aとおいて両辺をaで割ると求める式が得られます。