Question

求不定积分∫1/（4﹣x²）dx 麻烦写一下步骤·谢谢

Answer 1

根据高等数学基本积分表，∫1/(a^2-x^2)×dx=1/2a×ln[(a+x)/(a-x)]+C,所以…………

Answer 2

原式=-∫1/（x²-4）dx
=-1/4∫[1/（x-2）-1/（x+2)]dx
=-1/4∫1/(x-2)d(x-2)+1/4∫1/(x+2)d(x+2）
=-1/4ln|x-2|+1/4ln|x+2|
=-1/4ln|（x-2）/（x+2）|+c

Answer 3

∫1/（4﹣x²）dx=-1/4∫[1/(x+2)-1/(x-2)]dx=-1/4(lnΙx+2Ι-lnΙx-2Ι)+c

Answer 4

1/(4-x²)=1/(2-x)(2+x)=1/4[1/(2-x)+1/(2+x)]
∫1/(4-x²)dx
=1/4∫1/(2-x)+1/(2+x)dx
=1/4[-ln|2-x|+ln|2+x|]
=1/4ln(2+x)/(2-x)+C