Question

急，这题不定积分怎么求

Answer 1

=∫arctanxdx-∫x·arctanxdx

=x·arctanx-∫xdarctanx-½x²·arctanx+½∫x²darctanx

=(x-½x²)arctanx-∫xdx/(1+x²)+½∫x²dx/(1+x²)

=(x-½x²)arctanx-½∫d(1+x²)/(1+x²)+½∫(1+x²-1)dx/(1+x²)

=(x-½x²)arctanx-½ln(1+x²)+½x-arctanx+C

Answer 2

运用分部积分法
∫arctan(1–x)dx
=x·arctan(1–x)–∫xd[arctan(1–x)]
=x·arctan(1–x)＋∫x/(x²–2x＋2) dx
=x·arctan(1–x)＋∫(x–1)/(x²–2x＋2) dx＋∫1/[1²＋(x–1)²]dx
=x·arctan(1–x)＋ln∨(x²–2x＋2)＋arctan(x–1)＋C
=(x–1)·arctan(1–x)＋ln∨(x²–2x＋2)＋C