Question

∫(x+1)/(x²-2x+5)dx

Answer 1

∫[(x+1)/(x²-2x+5)]dx

=½∫[(2x-2+4)/(x²-2x+5)]dx

=½∫[(2x-2)/(x²-2x+5)]dx +2∫[1/(x²-2x+5)]dx

=½ln|x²-2x+5|+∫d(½x-½)/[1+(½x-½)²]

=½ln(x²-2x+5)+arctan(½x-½)+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

10、∫ secx dx =ln|cot(x/2)| + C 

= (1/2)ln|(1 + sinx)/(1 - sinx)| + C 

= - ln|secx - tanx| + C 

= ln|secx + tanx| + C

Answer 2

∫(x+1)/(x^2-2x+5) dx
=(1/2) ∫(2x-2)/(x^2-2x+5) dx +2∫dx/(x^2-2x+5)
=(1/2ln|x^2-2x+5| +2∫dx/(x^2-2x+5)
=(1/2ln|x^2-2x+5| +arctan[(x-1)/2]+C
consider
x^2-2x+5 =(x-1)^2 +4
let
x-1 = 2tanu
dx=2(secu)^2 du
∫dx/(x^2-2x+5)
=∫2(secu)^2 du/(4(secu)^2 )
=(1/2)∫ du
=(1/2)u+C1
=(1/2)arctan[(x-1)/2]+C1

Answer 3

∫[(x+1)/(x²-2x+5)]dx
=½∫[(2x-2+4)/(x²-2x+5)]dx
=½∫[(2x-2)/(x²-2x+5)]dx +2∫[1/(x²-2x+5)]dx
=½ln|x²-2x+5|+∫d(½x-½)/[1+(½x-½)²]
=½ln(x²-2x+5)+arctan(½x-½)+C