求不定积分∫(x^3)/(1+x^2)^3/2dx
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/08 09:04:39
求不定积分∫(x^3)/(1+x^2)^3/2dx
![求不定积分∫(x^3)/(1+x^2)^3/2dx](/uploads/image/z/19956664-64-4.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%28x%5E3%29%2F%281%2Bx%5E2%29%5E3%2F2dx)
∵∫[x^3/(1+x^2)^(3/2)]dx=(1/2)∫[(1+x^2-1)/(1+x^2)^(3/2)]d(1+x^2).
∴可令√(1+x^2)=y,得:1+x^2=y^2,∴d(1+x^2)=2ydy.
于是:
∫[x^3/(1+x^2)^(3/2)]dx
=(1/2)∫[(y^2-1)/y^3](2y)dy
=∫[(y^2-1)/y^2]dy
=∫dy-∫(1/y^2)dy
=y+1/y+C
=√(1+x^2)+1/√(1+x^2)+C.
∴可令√(1+x^2)=y,得:1+x^2=y^2,∴d(1+x^2)=2ydy.
于是:
∫[x^3/(1+x^2)^(3/2)]dx
=(1/2)∫[(y^2-1)/y^3](2y)dy
=∫[(y^2-1)/y^2]dy
=∫dy-∫(1/y^2)dy
=y+1/y+C
=√(1+x^2)+1/√(1+x^2)+C.