∫(x-1)/(x^2+2x+3)dx的不定积分怎么求
人气:226 ℃ 时间:2020-02-05 14:24:16
解答
∫(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
推荐
猜你喜欢
- Thanks for your bat.改为同意句
- 小石潭记中的文言现象
- ,一条路,甲单独修40天修完,乙单独修10天,再甲乙合作修20天修完,乙单独修多久修完 ,家乙合作多久修
- 加成反应的定义是什么?
- She ____(play) sports every day.横线上用 play 还是 plays啊?
- 我们经常在派对上玩得很开心,填英语.We often _____lots of_____ ______the party
- 设双曲线x^2/a^2-y^2/b^2=1半焦距为c,已知原点到bx+ay=ab的距离等于(1/4)c+1,则c最小为
- 写出一个大于六分之一,而小于五分之一的分数要过程