dx/(x^2+4x-5)的积分,怎么求?

题目

dx/(x^2+4x-5)的积分,怎么求?
请用把dx/(x^2+4x-5)变成dx/(（x+2）^2-9)接下去的方法做一遍,谢谢

答案

∫dx/(x²+4x-5),怎么求?【请用把∫dx/(x²+4x-5)变成∫dx/[(x+2)²-9]接下去的方法做一遍】
∫dx/(x²+4x-5)=∫dx/[(x+2)²-9]=∫dx/(x+2+3)(x+2-3)=∫dx/(x+5)(x-1)
=(1/6)∫[1/(x-1)-1/(x+5)]dx=(1/6)[∫d(x-1)/(x-1)-d(x+5)/(x+5)]=(1/6)[ln∣x-1∣-ln∣x+5∣]+C
=(1/6)ln∣(x-1)/(x+5)∣+C