Question

求一道微积分题的解答

∫dx/[x^根号下(a^+x^)]
^表示平方 根号下的是（a^+x^) 不好意思不会用搜狗打符号，这个题是在不会做，求不定积分的，谢谢高手指点
a>0

Answer 1

∫ dx/[x^2√(x^2+a^2)]
=∫ dx/[x^3√(1+(a/x)^2)]
=∫-1/2*d(1/x^2)*[1/√(1+(a/兄态消羡知x)^2)]
=∫-1/(2a^2)*d(a/x)^2*[1/√(1+(a/x)^2)]
=-1/(2a^2)∫闭团d[(a/x)^2+1]/√(1+(a/x)^2)]
= -1/a^2 *√[(a/x)^2+1]
= -1/a^2 * √(a^2+x^2)/x + C

令：x=atant , a/x = cott , sint=√csct^2=1/√[(a/x)^2+1]
∫ dx/[x^2√(x^2+a^2)]
=∫ asect^2 dt/[a^2tant^2 asect]
=∫ sect dt/[a^2tant^2]
=1/a^2∫ cost/ t^2 dt
=1/a^2∫ 1/sint^2 dsint
=1/a^2 [-1/sint] + C
= -1/a^2 * √(a^2+x^2)/x + C

Answer 2

你好
是亮橘这样的
∫ dx/[x^2√(x^2+a^2)]
=∫ dx/[x^3√雀消(1+(a/x)^2)]
=∫-1/2*d(1/x^2)*[1/√(1+(a/x)^2)]
=∫-1/(2a^2)*d(a/x)^2*[1/√(1+(a/x)^2)]
=-1/(2a^2)∫d[(a/x)^2+1]/√(1+(a/x)^2)]
= -1/a^2 *√[(a/x)^2+1]
= -1/a^2 * √顷键知(a^2+x^2)/x + C