作业帮求积分 dx/(x+根号1-x2)
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 01:36:09
求积分 dx/(x+根号1-x2)
求积分 dx/(x+根号1-x2)
求积分 dx/(x+根号1-x2)
令x=sint dx=costdt
原式=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost) B=∫sintdt/(sint+cost)
A+B=∫(sint+cost)dt/(sint+cost)=∫dt=t+C1
A-B=∫(cost-sint)dt/(sint+cost)=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C2
所以原式=A=(t+ln|sint+cost|)/2+C
=(arcsinx+ln|x+√(1-x^2)|)/2+C
原式=∫[1-√(1-x^2)]dx/x^2 //*分子分母同乘1-√(1-x^2),
设x=sint,dx=costdt,
(csct)^2=1/x^2,
(cott)^2=1/x^2-1=(1-x^2)/x^2.
cott=√(1-x^2)/x,
原式=∫(1-cost)*costdt/(sint)^2
=∫costdt/(sint)^2-∫(c...
全部展开
原式=∫[1-√(1-x^2)]dx/x^2 //*分子分母同乘1-√(1-x^2),
设x=sint,dx=costdt,
(csct)^2=1/x^2,
(cott)^2=1/x^2-1=(1-x^2)/x^2.
cott=√(1-x^2)/x,
原式=∫(1-cost)*costdt/(sint)^2
=∫costdt/(sint)^2-∫(cost/sint)^2dt
=∫d(sint)/(sint)^2-∫(cott)^2dt
=-1/(sint)-∫[csct)^2-1]dt
=-1/(sint)+cott+t+c
=-1/x+√(1-x^2)/x+arxsinx+C.
收起