Question

请问2项式如何展开，指数是负数的

比如（3x+2)^-2,有公式得告诉我，然后解一下

Answer 1

∫1/(3 x + 2)^2 d x
=-1/(3 (2 + 3 x))

∫-1/(3 (2 + 3 x)) d x
=-1/9 Log[6 + 9 x]

-1/9 Log[6 + 9 x]
展开为
-Log[6]/9 - x/6 + x^2/8 - x^3/8 + (9 x^4)/64 - (27 x^5)/160...

-Log[6]/9 - x/6 + x^2/8 - x^3/8 + (9 x^4)/64 - (27 x^5)/160...
对x求导得
-1/6 + x/4 - (3 x^2)/8 + (9 x^3)/16 - (27 x^4)/32...
再对小求导得
1/4 - (3 x)/4 + (27 x^2)/16 - (27 x^3)/8...

结果:
1/4 - (3 x)/4 + (27 x^2)/16 - (27 x^3)/8 + (405 x^4)/64 - (729 x^5)/64...

Answer 2

呃...我能想到的只有泰勒展开...