We are given the integral: \[ \int e^x \left( \frac{x + 5}{(x + 6)^2} \right) dx \] To solve this, we can use substitution. Let: \[ u = x + 6 \quad \Rightarrow \quad du = dx \] Thus, the integral becomes: \[ \int e^{u - 6} \left( \frac{u - 1}{u^2} \right) du \] Simplifying, we get: \[ e^{-6} \int e^u \left( \frac{u - 1}{u^2} \right) du \] By applying the method of integration by parts and solving, we end up with the result: \[ - \frac{e^x}{x + 6} \]
Thus, the correct answer is \( - \frac{e^x}{x + 6} \).
The integral $ \int_0^1 \frac{1}{2 + \sqrt{2e}} \, dx $ is:
If a random variable X has the following probability distribution values:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|---|
P(X) | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 |
Then P(X ≥ 6) has the value: