To find the coefficient of x^5 in the expansion of (1 + x^2)^5(1 + x)^4, we can use the binomial theorem.

The binomial theorem states that for any real numbers a and b, and any positive integer n, the expansion of (a + b)^n can be written as the sum of the terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient.

In this case, we have (1 + x^2)^5(1 + x)^4. We can expand this using the binomial theorem:

(1 + x^2)^5(1 + x)^4 = C(5, 0) * 1^5 * (x^2)^0 * 1^4 + C(5, 1) * 1^4 * (x^2)^1 * 1^3 + C(5, 2) * 1^3 * (x^2)^2 * 1^2 + C(5, 3) * 1^2 * (x^2)^3 * 1^1 + C(5, 4) * 1^1 * (x^2)^4 * 1^0 + C(5, 5) * 1^0 * (x^2)^5 * 1^0

Now, let's look at the term that contains x^5. We can see that the only way to obtain x^5 is by multiplying (x^2)^2 and (x^2)^3:

C(5, 2) * 1^3 * (x^2)^2 * 1^2 * C(5, 3) * 1^2 * (x^2)^3 * 1^1

Simplifying this expression, we get:

C(5, 2) * C(5, 3) * x^4 * x^6

The binomial coefficient C(5, 2) represents the number of ways to choose 2 items from a set of 5, which is equal to 10. Similarly, C(5, 3) represents the number of ways to choose 3 items from a set of 5, which is also equal to 10. Therefore, the coefficient of x^5 is:

10 * 10 * x^4 * x^6 = 100 * x^4 * x^6 = 100x^10

So, the coefficient of x^5 in the expansion of (1 + x^2)^5(1 + x)^4 is 0.