Question

求（x－1）／（x^2+2x+3）的不定积分

要详细过程，谢谢

Answer 1

∫x3/(x2+2x-3)dx=∫(x3+2x-3x-2x+3)/(x2+2x-3)dx =∫x+3/(x2+2x-3)dx =∫xdx+3∫1/(x2+2x-3)dx =x2/2+3∫1/[(x-1)(x+1)]dx =x2/2+3/4∫1/(x-1)-1/(x+3)]dx = x2/2+3/4ln|x-1|-3/4ln|x+3|+C

Answer 2

1/(x^2-2x-3) = (1/4)[1/(x-3) -1/(x+1)]

∫dx/(x^2-2x-3)
=(1/4)∫[1/(x-3) -1/(x+1)] dx
=(1/4) ln|(x-3)/(x+1)| + C

Answer 3

我想问一下第三步的后面一部分怎么解的

Answer 4

原式＝∫（x+1-2）/（x²+2x+3）dx
＝∫（x²/2+x）/（x²+2x+3）dx-∫2/[2+（1+x）²]dx
＝1/2*ln（x²+2x+3）-∫1/[1+（1/✓2 +x/✓2）²]dx
＝1/2*ln（x²+2x+3）-✓2*arctan（x/✓2+1/✓2）+C