Question

不定积分∫(1/根号下x^2+4的平方根)dx？

Answer 1

let x=2tanu dx=2(secu)^2 du ∫ dx/√(x^2+4)=∫ 2(secu)^2 du/(2secu)=∫secu du =ln|secu +tanu| +C'=ln| √(x^2+4)/2+x/2| +C'=ln| √(x^2+4)+x| +C ...

Answer 2

let
x=2tanu
dx=2(secu)^2 du
∫ dx/√(x^2+4)
=∫ 2(secu)^2 du/(2secu)
=∫secu du
=ln|secu +tanu| +C'
=ln| √(x^2+4)/2+x/2| +C'
=ln| √(x^2+4)+x| +C

Answer 3

令 t=9-x² => dt=-2xdx∫x³/√(9-x²) dx= ∫(9-t)x/√t * -1/(2x) dt= (-1/2)∫( 9-t)/√t dt= (-1/2)∫(9/√t - √t) dt= (1/2)∫√t dt - (9/2)∫1/√t dt= ( 1/3)t^(3/2) - 9√t + C= (1/3)(9-x²)^(3/2) - 9√(9-x²) + C= (-1/3 )(x²+18)√(9-x²) + C