x⼀(x^2+2x-3)^1⼀2的不定积分

2024-11-08 11:13:22
推荐回答(2个)
回答(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

回答(2):

∫ x/√(x² + 2x - 3) dx
= ∫ x/√[(x + 1)² - 4] dx,x + 1 = 2secz,dx = 2secztanz dz
= ∫ (2secz - 1)/|2tanz| * (2secztanz dz)
= ∫ (2secz - 1) * secz dz
= 2∫ (sec²z - secz) dz
= 2tanz - ln|secz + tanz| + C
= √(x² + 2x - 3) - ln|x + 1 + √(x² + 2x - 3)| + C
————————————————————
secz = (x + 1)/2
tanz = √(sec²z - 1) = √[((x + 1)/2)² - 1] = √[(x² + 2x + 1) - 4]/2 = √(x² + 2x - 3)/2