Question

高数，不定积分，求解释，求过程

Answer 1

Answer 2

f(x)的不定积分为J f(x) dx
1/[(x+1)^2 (x^2 + 1)] = 1/[2x (x^2 + 1)] - 1/[2x (x + 1)^2]
1/[2x (x^2 + 1)] = x/2 * /[x^2 (x^2+1)] =x/2 * [1/x^2 - 1/(x^2 + 1)]
1/[2x (x + 1)^2] = 1/2 * [1/(x *(x+1)) - 1/(x+1)^2] = 1/2 * [1/x - 1/(x+1) - 1/(x+1)^2]

J 1/[2x (x^2 + 1)]dx
= J x/2 * [1/x^2 - 1/(x^2 + 1)] dx
= J 1/4 * [1/x^2 - 1/(x^2 + 1)] dx^2
= 1/4 * [J 1/x^2 dx^2 - J 1/(x^2 + 1) dx^2]
= 1/4 * [J 1/x^2 dx^2 - J 1/(x^2 + 1) d(x^2+1)]
= 1/4 * [ln(x^2) - ln(x*2 +1)]
= 1/4 * ln(x^2/(x^2 + 1))
J 1/[2x (x + 1)^2] dx
=J 1/2 * [1/x - 1/(x+1) - 1/(x+1)^2] dx
=1/2 * (J 1/x dx - J 1/(x+1) dx - J 1/(x+1)^2 dx
= 1/2 (ln(x/(x+1)) + 1/(x+1))
带进去就可求出。