Question

Find the integral
∫√(9+x)/(9-x)

Answer 1

I suppose you mean

`\displaystyle \int (√(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 (2y^2)/(18 - y^2) \, dy`

Simplify the integrand by dividing:

`(2y^2)/(18 - y^2) = -2 + (36)/(18 - y^2)`

`\implies \displaystyle \int \left((36)/(18-y^2) - 2\right) \, dy`

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

`\frac1{18 - y^2} = \frac1{√(72)} \left(\frac1{√(18)-y} + \frac1{√(18)+y}\right)`

`\implies \displaystyle (36)/(√(72)) \int \left(\frac1{√(18)-y} + \frac1{√(18)+y}\right) \, dy - 2 \int dy`

The rest is trivial:

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

`= \displaystyle √(18) \left(\ln\left|√(18)+y\right| - \ln\left|√(18)-y\right|\right) - 2y + C`

`= \displaystyle √(18) \ln\left|(√(18)+y)/(√(18)-y)\right| - 2y + C`

`= \boxed - 2√(9+x) + C`