Question

Find the integral ∫√(9+x)/(9-x)

Answer 1

I suppose you mean

$\displaystyle \int \frac{\sqrt{9+x}}{9-x} \, dx$

Substitute y = √(9 + x). Solving for x gives x = y² - 9, so that 9 - x = 18 - y², and we have differential dx = 2y dy. Replacing everything in the integral gives

$\displaystyle \int \frac{2y^2}{18 - y^2} \, dy$

Simplify the integrand by dividing:

$\dfrac{2y^2}{18 - y^2} = -2 + \dfrac{36}{18 - y^2}$

$\implies \displaystyle \int \left(\frac{36}{18-y^2} - 2\right) \, dy$

For the first term of this new integral, we have the partial fraction expansion

$\dfrac1{18 - y^2} = \dfrac1{\sqrt{72}} \left(\dfrac1{\sqrt{18}-y} + \dfrac1{\sqrt{18}+y}\right)$

$\implies \displaystyle \frac{36}{\sqrt{72}} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

The rest is trivial:

$\displaystyle \sqrt{18} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

$= \displaystyle \sqrt{18} \left(\ln\left|\sqrt{18}+y\right| - \ln\left|\sqrt{18}-y\right|\right) - 2y + C$

$= \displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+y}{\sqrt{18}-y}\right| - 2y + C$

$= \boxed{\displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+\sqrt{9+x}}{\sqrt{18}-\sqrt{9+x}}\right| - 2\sqrt{9+x} + C}$