高数 ∫3x/x^2+2x+17dx积分如何解?

用手机不太好答…
350200039答得很不错,至少方法是对了,可惜犯了一些很低级的错误...
3 ∫(x+1-1)/[(x+1)^2+16]dx=
3 ∫0.5/[(x+1)^2+16]d[(x+1)^2+16]-3∫[1/[(x+1)^2+16]d(x+1)
=3/2ln[(x+1)^2+16]-3/16∫[1/{[(x+1)/4]^2+1}d(x+1)
=3/2ln[(x+1)^2+16]-3/4∫[1/{[(x+1)/4]^2+1}d[(x+1)/4]
=3/2ln[(x+1)^2+16]-3/4arctan[(x+1)/4]+C

我认为你想表达的是3x/(x^2+2x+17)吧
变为3 ∫(x+1-1)/[(x+1)^2+16]dx=
3 ∫0.5/[(x+1)^2+16]d[(x+1)^2+1]-3∫(1/[(x+1)^2+16]d(x+1)
=3/2ln[(x+1)^2+1]-3arctan (x+1) +c

∫3x/x^2+2x+17dx=∫3/x+2x+17dx=3LnX+X^2+17X+C（C代表常数）