Question

求它的原函数

Answer 1

直接查表啊

Answer 2

∫ (x^2+a^2)^(3/2) dx
=x(x^2+a^2)^(3/2) - (4/3)∫ x^2. (x^2+a^2)^(1/2) dx
=x(x^2+a^2)^(3/2) - (4/3)∫ (x^2+a^2)^(3/2) dx +(4/3)a^2∫ (x^2+a^2)^(1/2) dx
(7/3)∫ (x^2+a^2)^(3/2) dx =x(x^2+a^2)^(3/2) +(4/3)a^2∫ (x^2+a^2)^(1/2) dx
∫ (x^2+a^2)^(3/2) dx
=(3/7) { x(x^2+a^2)^(3/2) +(2/3)a^4. [x √(x^2+a^2) /a^2 + ln|√(x^2+a^2)+ x| ] }+ C
-----------------------
let
x= atanu
dx = a(secu)^2 du
∫ (x^2+a^2)^(1/2) dx
=∫ (asecu)( a(secu)^2 du )
=a^2 ∫ (secu)^3 du
=(1/2)a^2. [secu.tanu + ln|secu+tanu| ] + C''
=(1/2)a^2. [x √(x^2+a^2) /a^2 + ln|√(x^2+a^2) /a+ x/a| ] + C''
=(1/2)a^2. [x √(x^2+a^2) /a^2 + ln|√(x^2+a^2)+ x| ] + C'''
consider
∫ (secu)^3 du = ∫ secu dtanu
=secu.tanu - ∫ (tanu)^2. secu du
=secu.tanu - ∫ [(secu)^2-1]. secu du
2∫ (secu)^3 du =secu.tanu + ∫ secu du
∫ (secu)^3 du =(1/2)[secu.tanu + ln|secu+tanu| ] +C'