作业帮求积分,被积函数是ln[(x+√(x²+1))/(x+√(x²-1))],
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![求积分,被积函数是ln[(x+√(x²+1))/(x+√(x²-1))],](/uploads/image/z/6096806-62-6.jpg?t=%E6%B1%82%E7%A7%AF%E5%88%86%2C%E8%A2%AB%E7%A7%AF%E5%87%BD%E6%95%B0%E6%98%AFln%5B%28x%2B%E2%88%9A%28x%26%23178%3B%2B1%29%29%2F%28x%2B%E2%88%9A%28x%26%23178%3B-1%29%29%5D%2C)
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求积分,被积函数是ln[(x+√(x²+1))/(x+√(x²-1))],
求积分,被积函数是ln[(x+√(x²+1))/(x+√(x²-1))],
求积分,被积函数是ln[(x+√(x²+1))/(x+√(x²-1))],
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
原式=ln[(x+√(x²+1)]-ln[x+√(x²-1))]原式'=[1+2x/√(x²+1)]/[(x+√(x²+1)]-[1+2x/√(x²-1)]/[(x+√(x²-1)]