Find $$ \int \frac{2 x+6}{\left(x^{2}+6 x-9\right)^{5}} d x $$ Answer: $$ \square $$

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Calculate the integral $$ \frac{2x+6}{(x^2+6x-9)^5} $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \frac{2x+6}{\left(x^{2}+6x-9\right)^{5}} dx$$
- step1: Use the substitution $$dx=\frac{1}{2x+6} dt$$ to transform the integral$$:$$
  $$\int \frac{2x+6}{\left(x^{2}+6x-9\right)^{5}}\times \frac{1}{2x+6} dt$$
- step2: Simplify:
  $$\int \frac{1}{\left(x^{2}+6x-9\right)^{5}} dt$$
- step3: Use the substitution $$t=x^{2}+6x-9$$ to transform the integral$$:$$
  $$\int \frac{1}{t^{5}} dt$$
- step4: Evaluate the integral:
  $$\frac{t^{-5+1}}{-5+1}$$
- step5: Add the numbers:
  $$\frac{t^{-4}}{-5+1}$$
- step6: Add the numbers:
  $$\frac{t^{-4}}{-4}$$
- step7: Substitute back:
  $$\frac{\left(x^{2}+6x-9\right)^{-4}}{-4}$$
- step8: Simplify:
  $$\frac{\frac{1}{\left(x^{2}+6x-9\right)^{4}}}{-4}$$
- step9: Add the constant of integral C:
  $$\frac{\frac{1}{\left(x^{2}+6x-9\right)^{4}}}{-4} + C, C \in \mathbb{R}$$
The integral $$ \int \frac{2x+6}{(x^2+6x-9)^5} dx $$ can be calculated using partial fraction decomposition. The result is:

$$ \frac{1}{-4(x^2+6x-9)^4} + C $$

where $$ C $$ is an arbitrary constant.
$$ \frac{1}{-4(x^2+6x-9)^4} + C $$

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