Question

高数不定积分题

∫dx/(x²+a²)^n=x/[2(n-1)a²(x²+a²)^n-1]+[(2n-3)/(2(n-1)a²]∫dx/(x²+a²)^n-1

Answer 1

像∫dx/(x²+a²)^n的积分要用递推关系求解
设In=∫dx/(x²+a²)²
则I(n+1)=x/[2na²(x²+a²)^n]+(2n-1)/2na²In
而I1=∫dx/(x²+a²)=1/aarctan(x/a)+c

Answer 2

∫dx/(x^2+a^2)^(n-1)=x/(x^2+a^2)^(n-1)-∫xd(1/(x^2+a^2)^(n-1))
=x/(x^2+a^2)^(n-1)+(n-1)∫2x^2dx/(x^2+a^2)/x^n
=x/(x^2+a^2)^(n-1)+2(n-1)∫dx/(x^2+a^2)^(n-1) -2(n-1)∫a^2dx/(x^2+a^2)^n
(3-2n)∫dx/(x^2+a^2)^(n-1)=x/(x^2+a^2)^(n-1) -2(n-1)a^2∫dx/(x^2+a^2)^n
(2n-3)/((2n-1)a^2)∫dx/(x^2+a^2)^(n-1)=∫dx/(x^2+a^2)^n- (1/(2(n-1)a^2))*x/(x^2+a^2)^(n-1)

∫dx/(x^2+a^2)^n= (1/(2(n-1)a^2))*x/(x^2+a^2)^(n-1)+∫(2n-3)/((2n-1)a^2)∫dx/(x^2+a^2)^(n-1)

Answer 3

你是要求什么东西？