求不定积分 ∫8/x^2-4 dx求详细过程
问题描述：
求不定积分 ∫8/x^2-4 dx
求详细过程
答
原式=∫8/((x+2)(x-2))dx=∫2(1/(x-2)-1/(x+2))dx=2∫dx/(x-2)-2∫dx/(x+2)=2ln|x-2|-2ln|x+2|+C=ln((x-2)/(x+2))^2+C
答
∫8/x^2-4 dx
=8∫dx/(x+2)(x-2)
=8∫dx/[1/(x-2)-1/(x+2)]*1/4
=2∫dx/(x-2)-2∫dx/(x+2)
=2ln|x-2|-2ln|x+2|+C
=ln(x-2)^2-ln(x+2)^2+C
=ln[(x-2)/(x+2]^2 +C