Question

求不定积分1/x^2+2x+3

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

令u=x+1
则du=dx
原式=∫1/(2+u²2)du
=1/√2·arctan(u/√2)+C
=1/√2·arctan[(x+1)/√2]+C

追答

令u=x+1
则du=dx
原式=∫1/(2+u^2)du
=1/√2·arctan(u/√2)+C
=1/√2·arctan[(x+1)/√2]+C

追问

为什么可以写1/根号2啊！怎么来的

追答

∫1/(a^2+x^2)dx
=1/a·arctan(x/a)+C

Answer 3