Use the substitution u=3+(x+4)^1/2 to find the integral of 1/(3+(x+4)^1/2) dx between 0 and 5.

We will call the integral I, so I = integral of 1/(3+(x+4)^1/2) dx between 0 and 5. First substitute u=3+(x+4)^1/2 into the equation to get I = integral of 1/u dx between 0 and 5 Next we want to change the 'dx' into a 'du' as our integral now needs to be with respect to u as that is the variable in the integral. We do this by finding dx in terms of du by differentiating u with respect to x: du/dx = (1/2)*(x+4)^-1/2 dx = 2(x+4)^1/2 du = 2(u-3)     Substituting this into the equation gives you: I = integral of 2(u-3)/u du  = integral of 2-(6/u)    However this can no longer be between 0 and 5 as those were the values x would have taken. Now that the integral is with respect to u, we need to find the values that u would take. If x=5, u=3+(5+4)^1/2=6 If x=0, u=3+(0+4)^1/2=5 So now I = integral of 2-(6/u) du between 5 and 6 This gives [2u-6ln(u)] between 5 and 6 = (12-6ln(6))-(10-6ln(5)) = 2 + 6ln(5) - 6ln(6) = 2 +6ln(5/6)