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  • Quirkie
    What to do you need to do with that? Just for fun, I'll integrate it. It's in two parts: sqrt(x+3)/x and 2/x the second bit is 2 log x For the first bit I put u = sqrt(x+3) x = u^2 - 3 du/dx = (1/2) / u so dx = 2 u du Now it is integral 2 u^2 / (u^2 - 3) du by partial fractions: = 2 + 6/(u^2 - 3) du The first part integrates to 2 u The second part: hm. tanh^2 - 1 is sinh^2/cosh^2 - cosh^2/cosh^2 = (sinh^2 - cosh^2) / cosh^2 = -1/cosh^2 so put u=sqrt(3) tanh k u^2 - 3 = -3 ( 1/cosh^2) du/dk = sqrt(3) (1/cosh^2) integral (6/(u^2 - 3)) du = integral 6 sqrt(3) / -3 dk = - (2 sqrt(3)) k
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