其实这个可以直接分部积分,但后边求导起来比较复杂.那就间接分部积分,不过结果是一样的
∫ln(x+√(x²+1)-ln(x+√(x²-1))dx
=∫ln(x+√(x²+1)dx-∫ln(x+√(x²-1))dx
=xln(x+√(x²+1)-∫xd[ln(x+√(x²+1)]-xln(x+√(x²-1))+∫xd[ln(x+√(x²-1))] //这里的求导仔细点就可以了,我相信你会求的.
=x[ln(x+√(x²+1)-ln(x+√(x²-1))]-∫x/√(x^2+1)dx+∫x/√(x^2-1) dx
=xln[(x+√(x²+1))/(x+√(x²-1))]+√(x^2-1)-√(x^2+1)+C