c) ∫(x + 4)/(x² + 8x - 9)dx

Explanation:

Step1: Let (u = x^{2}+8x - 9)

Differentiate (u) with respect to (x): (du=(2x + 8)dx=2(x + 4)dx), so ((x + 4)dx=\frac{1}{2}du)

Step2: Substitute into the integral

The integral (\int\frac{x + 4}{x^{2}+8x - 9}dx) becomes (\int\frac{\frac{1}{2}du}{u}=\frac{1}{2}\int\frac{du}{u})

Step3: Integrate (\frac{1}{u})

We know that (\int\frac{du}{u}=\ln|u|+C)

Step4: Substitute (u) back

Substituting (u=x^{2}+8x - 9) back, we get (\frac{1}{2}\ln|x^{2}+8x - 9|+C)

Answer:

(\frac{1}{2}\ln|x^{2}+8x - 9|+C)