Question

求（x－1）／（x^2+2x+3）的不定积分

要详细过程，谢谢

Answer 1

∫(x-1)/(x²+2x+3)dx

=½∫(2x-2)/(x²+2x+3)dx

=½∫(2x+2-4)/(x²+2x+3)dx

=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx

=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx

=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx

=½ln|x²+2x+3| - ∫1/{[(x+1)/√2]²+1}dx + C

=½ln|x²+2x+3| - (√2)∫1/{[(x+1)/√2]²+1}d[(x+1)/√2] + C

=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + C

扩展资料

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

9、∫ tanx dx = - ln|cosx| + C = ln|secx| + C

Answer 2

详细过程如图所示，令x+1=t换元做，希望对你有所帮助，望采纳哦

Answer 3

Answer 4

令w=x^1/6
则x=w^6,dx=6w^5dw
则原式=6∫w^3/(w+1)dw=6∫(w^3+1-1)/(w+1)dw
=6∫[(w^2-w+1)-1/(w+1)]dw=2w^3-3w^2+6w-ln(w+1)+C
带入w=x^1/6
得原式=2x^1/2-3x^1/3+6x^1/6-ln(1+x^1/6)+C
楼上的代换形式也是正确的，但在中间计算过程中可能有错误。

Answer 5

∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx
=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx
=½ln|x²+2x+3| - ∫1/{[(x+1)/√2]²+1}dx + C
=½ln|x²+2x+3| - (√2)∫1/{[(x+1)/√2]²+1}d[(x+1)/√2] + C
=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + C