Evaluate
4 √(ln(9-x)) dx
∫ ------------------------
2 √(ln(9-x)) + √(ln(x+3))
There is a trick that can be used for definite integrals that otherwise look hard. If you already know this trick, my solution will seem rather wordy.
We are going to make a substitution based on the bounds of integration: u = 6-x. (Then dx = -du)
Starting integral
I = Integ {2 to 4} sqrt(ln(9-x))/[sqrt(ln(9-x))+sqrt(ln(3+x))] dx
Applying the substitution gives us
I = Integ {4 to 2} -sqrt(ln(3+u))/[sqrt(ln(3+u))+sqrt(ln(9-u))] dx
Now flip the bounds of integration to cancel the minus sign.
I = Integ {2 to 4} sqrt(ln(3+u))/[sqrt(ln(3+u))+sqrt(ln(9-u))] dx
Now the first and third forms of the integral have the same bounds, but differ only in the variable, but since this is a definite integral we can just make the trivial substitution u=x.
Then add those two forms of I together to get
2*I = Integ {2 to 4} [sqrt(ln(3+u))+sqrt(ln(9-u))]/[sqrt(ln(3+u))+sqrt(ln(9-u))] dx
Now this is convenient, the integrand simplifies down to just 1. Then all we have left is the trivial integral
2*I = Integ {2 to 4} du
2*I = 4 - 2
I = 1
The value of the integral is 1.
Solution
