Let $y(x) = a(x + 1) \log(x + 1) + bx + 5$ be the solution of the differential equation e$^\frac{dy}{dx} = x + 1_{;}y(0) = 5$, then the value of $(a + b)$ is:

We need to find $y(x) = a(x+1)\log(x+1) + bx + 5$ that satisfies $e^{\frac{dy}{dx}} = x+1$ with $y(0) = 5$, and then find $(a+b)$.

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Taking natural logarithm on both sides of $e^{\frac{dy}{dx}} = x+1$:

$\dfrac{dy}{dx} = \ln(x+1)$

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Now, differentiating the given $y(x) = a(x+1)\ln(x+1) + bx + 5$ using the product rule on the first term:

$\dfrac{dy}{dx} = a\left[\ln(x+1) + 1\right] + b$

$= a\ln(x+1) + a + b$

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Setting this equal to what we found from the differential equation:

$a\ln(x+1) + (a + b) = \ln(x+1)$

Comparing both sides:

Coefficient of $\ln(x+1)$: $a = 1$

Constant term: $a + b = 0 \Rightarrow b = -1$

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Verifying the initial condition $y(0) = 5$:

$y(0) = 1 \cdot (1)\ln(1) + (-1)(0) + 5 = 0 + 0 + 5 = 5$ ✓

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$a + b = 0$