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make a substitution to express the integrand as a rational function and then evaluate the integral

Make the substitution z = (x+3)^1/2 so that z^2 = x+3 (or x = z^2-3) and dx = 2zdz then

dx/[2(x+3)^1/2+x] = 2zdz/[z^2+2z-3] = 2zdz/[ (z-1)(z+3) ] <= rational function

Using partial fractions, set

2z/[ (z-1)(z+3) ] = A/(z-1) + B/(z+3) where A and B are undetermined coefficients. Multiply each side by (z-1)(z+3) and match powers of z to get

A+B = 2 and 3A-B = 0 so that A = 3/2 and B = 1/2. The integral becomes

(1/2)∫dz/(z-1) + (3/2)∫dz/(z+3) which is readily integrated;

(1/2)ln(z-1) + (3/2)ln(z+3)

back substituting z = (x+3)^1/2

∫ = (1/2)ln[(x+3)^1/2-1] + (3/2)ln[(x+3)+3] + const

Further simplifications are possible (eg., a･lnb = ln(b^a) etc.) and you might also look more carefully at (ąz)^2 = x+3 (I chose the plus sign).

You need to check my algebra but this is basically how to do the integral.

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