Equate the tangent line to the graph of the function ln (x ^ 2 + 2x) at the point with the abscissa x = 2.

1. Let’s find the derivative of the function:

y = ln (x ^ 2 + 2x);

y ‘(x) = (x ^ 2 + 2x)’ / (x ^ 2 + 2x);
y ‘(x) = (2x + 2) / (x (x + 2));
y ‘(x) = 2 (x + 1) / (x (x + 2)).
2. Let’s calculate the values of the function and the derivative at the point with the abscissa x0 = 2:

y0 = y (2) = ln (2 ^ 2 + 2 * 2) = ln (4 + 4) = ln8 = 3ln2;
y ‘(2) = 2 (2 + 1) / (2 (2 + 2)) = 2 * 3/8 = 3/4.
3. The equation of the tangent to the graph of the function at the point (x0; y0):

(y – y0) / (x – x0) = y ‘(y0);

(y – 3ln2) / (x – 2) = 3/4;

y – 3ln2 = (3/4) (x – 2);

y = 3ln2 + 3x / 4 – 3/2;

y = 3x / 4 + 3ln2 – 3/2.

Answer: y = 3x / 4 + 3ln2 – 3/2.