Answer is: option4

Solution:

We are given the integral:

\[ \int \frac{16}{x^2 - 6x + 25} \, dx \]

The quadratic in the denominator is:

\[ x^2 - 6x + 25 = (x - 3)^2 + 16 \]

So we rewrite the integral as:

\[ \int \frac{16}{(x - 3)^2 + 4^2} \, dx \]

Let:

\[ u = \frac{x - 3}{4} \Rightarrow x - 3 = 4u \Rightarrow dx = 4 \, du \]

Substitute into the integral:

\[ \int \frac{16}{(x - 3)^2 + 16} \, dx = \int \frac{16}{(4u)^2 + 4^2} \cdot 4 \, du = \int \frac{64}{16(u^2 + 1)} \, du = 4 \int \frac{1}{u^2 + 1} \, du \]

We know:

\[ 4 \int \frac{1}{u^2 + 1} \, du = 4 \tan^{-1}(u) + C \]

Substitute back \( u = \frac{x - 3}{4} \):

\[ = 4 \tan^{-1} \left( \frac{x - 3}{4} \right) + C \]

Hence option D