求不定积分(x^2-a^2)^3/2a是常数

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 14:47:45

求不定积分(x^2-a^2)^3/2a是常数
求不定积分(x^2-a^2)^3/2
a是常数

求不定积分(x^2-a^2)^3/2a是常数
答:


∫ (x²-a²)^(3/2) dx
=∫ (x²-a²)√(x²-a²) dx
=∫ x²√(x²-a²) dx -a² ∫√(x²-a²) dx
=x(x²-a²)^(3/2) - ∫ x*(3/2)*(2x)*(x²-a²)^(1/2) dx
=x(x²-a²)^(3/2) - 3 ∫ (x²)√(x²-a²) dx
所以:
4∫ (x²)√(x²-a²) 
=x(x²-a²)^(3/2)+a²∫√(x²-a²) dx
=x(x²-a²)^(3/2)+(a²x /2)√(x²-a²)-(1/2)*(a^4)*ln|x+√(x²-a²)|


所以:
∫ (x²)√(x²-a²) =(1/4)x(x²-a²)^(3/2)+(1/8)*(a²x)√(x²-a²)-(1/8)*(a^4)*ln|x+√(x²-a²)|
所以:

∫ (x²-a²)^(3/2) dx
=x(x²-a²)^(3/2) - 3 ∫ (x²)√(x²-a²) dx
=x(x²-a²)^(3/2) - (3/4)x(x²-a²)^(3/2)-(3/8)*(a²x)√(x²-a²)+(3/8)*(a^4)*ln|x+√(x²-a²)|+C
=(1/4)x(x²-a²)^(3/2)-(3/8)*(a²x)√(x²-a²)+(3/8)*(a^4)*ln|x+√(x²-a²)|+C


a是常数?