求不定积分:1、∫1/[x^2(x^2+1)]dx 2、∫sinx/(1+sinx)dx
人气:288 ℃ 时间:2019-08-19 12:13:40
解答
1、∫1/[x²(x²+1)]dx
=∫[1/x²-1/(x²+1)]dx
=∫dx/x²-∫dx/(x²+1)
=-1/x-arctanx+C (C是积分常数)
2、∫sinx/(1+sinx)dx
=∫(1+sinx-1)/(1+sinx)dx
=∫[1-1/(1+sinx)]dx
=∫dx-∫dx/(1+sinx)
=x-∫dx/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]
=x-∫dx/[sin(x/2)+cos(x/2)]²
=x-∫sec²(x/2)/[tan(x/2)+1]²dx
=x-∫d[tan(x/2)]/[tan(x/2)+1]²dx
=x-∫d[tan(x/2)+1]/[tan(x/2)+1]²dx
=x+1/[tan(x/2)+1]+C (C是积分常数)
推荐
猜你喜欢
- 将军既帝室之胄
- 我的理想作文400字
- 如果15分之a是最简真分数,那么a可取的整数有( )个
- _______ you can't answer this question,we have to ask someone else.A.While B.Before C.When D.since
- 从集合{-2,-1,0,1,2,3}中,任取三个元素作为二次函数y=ax的平方+bx+c的系数的值,则所得抛物线(即抛
- 弹簧弹力做正功时,弹性势能一定减少?
- 关于‘心’字成语
- 甲乙两人沿同一条路骑自行车(匀速)从A站到B站,甲需要30分钟,乙需要40分钟,如果乙比甲早出发5分钟去B站,则甲出发后经_分钟可以追上乙.