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\( \int \frac { 1 + \frac { d y } { d x } } { ( x + y ) ^ { 2 } } \)

Ask by Lindsey Delgado. in Bangladesh
Feb 25,2025

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Tutor-Verified Answer

Answer

The integral simplifies to \( -\frac{1}{x + y} + C \).

Solution

We start with the integral   ∫ [1 + (dy/dx)] / (x + y)² dx. Notice that the numerator is exactly the derivative of (x + y) with respect to x. That is,   d/dx (x + y) = 1 + (dy/dx). This observation suggests using the substitution   u = x + y   ⇒  du/dx = 1 + (dy/dx),   or equivalently, du = [1 + (dy/dx)] dx. With this substitution, the integral becomes   ∫ du/u². Now, recalling the standard antiderivative:   ∫ u^(–2) du = -u^(–1) + C, we substitute back to obtain   -1/(x + y) + C. Thus, the integral evaluates to   -1/(x + y) + C.

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Bonus Knowledge

To approach the integral \( I = \int \frac { 1 + \frac { d y } { d x } } { ( x + y ) ^ { 2 } } \), we can split the integral into two parts for easier calculation: 1. The first part is \( \int \frac{1}{(x+y)^2} \, dx \). 2. The second part involves \( \int \frac{\frac{dy}{dx}}{(x+y)^2} \, dx \). For the first part, you can use a simple substitution if necessary, or process it directly based on the form. The second part requires integration by recognizing that \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \), which can be tricky if we don't know the relationship between \( x \) and \( y \). From here, you can evaluate each integral based on the specific relationship between \( x \) and \( y \) intended in your problem!

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