demo

Integral Solution We start with the integral: \[ I = \int \frac{dx}{(x+2)\sqrt{x+11}} \] Let \( u = \sqrt{x + 11} \). Then: \[ u^2 = x + 11 \implies x = u^2 - 11 \quad \text{and} \quad dx = 2u \, du \] Rewrite the Integral Substitute \( x + 2 = u^2 - 9 \) into the original integral: \[ I = \int \frac{2u \, du}{(u^2 - 9)u} = 2 \int \frac{du}{u^2 - 9} \] Partial Fraction Decomposition Recognize the standard integral form: \[ \int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln\left|\frac{u - a}{u + a}\right| + C \] For \( a = 3 \): \[ 2 \int \frac{du}{u^2 - 9} = 2 \cdot \frac{1}{6} \ln\left|\frac{u - 3}{u + 3}\right| + C = \frac{1}{3} \ln\left|\frac{u - 3}{u + 3}\right| + C \] Substitute Back Replace \( u \) with \( \sqrt{x + 11} \): \[ \frac{1}{3} \ln\left|\frac{\sqrt{x + 11} - 3}{\sqrt{x + 11} + 3}\right| + C \] Final Answer \[ \boxed{\frac{1}{3} \ln\left|\frac{\sqrt{x + 11} - 3}{\sqrt{x + 11} + 3}\right| + C} \]

Question2

\[ u = \log \left( \frac{x^2 + y^2}{x + y} \right) \] We need to find \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \).

solution

: First, we compute the partial derivative with respect to \( x \): \[ \frac{\partial u}{\partial x} = \frac{1}{\frac{x^2 + y^2}{x + y}} \cdot \frac{\partial}{\partial x} \left( \frac{x^2 + y^2}{x + y} \right) \] Using the quotient rule: \[ \frac{\partial}{\partial x} \left( \frac{x^2 + y^2}{x + y} \right) = \frac{(x + y) \frac{\partial}{\partial x}(x^2 + y^2) - (x^2 + y^2) \frac{\partial}{\partial x}(x + y)}{(x + y)^2} \] \[ = \frac{(x + y)(2x) - (x^2 + y^2)(1)}{(x + y)^2} \] \[ = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2} \] \[ = \frac{2x^2 + 2xy - x^2 - y^2}{(x + y)^2} \] \[ = \frac{x^2 + 2xy - y^2}{(x + y)^2} \] Substitute back: \[ \frac{\partial u}{\partial x} = \frac{x + y}{x^2 + y^2} \cdot \frac{x^2 + 2xy - y^2}{(x + y)^2} \] \[ = \frac{x^2 + 2xy - y^2}{(x^2 + y^2)(x + y)} \] Next, we compute the partial derivative with respect to \( y \): \[ \frac{\partial u}{\partial y} = \frac{1}{\frac{x^2 + y^2}{x + y}} \cdot \frac{\partial}{\partial y} \left( \frac{x^2 + y^2}{x + y} \right) \] Using the quotient rule: \[ \frac{\partial}{\partial y} \left( \frac{x^2 + y^2}{x + y} \right) = \frac{(x + y) \frac{\partial}{\partial y}(x^2 + y^2) - (x^2 + y^2) \frac{\partial}{\partial y}(x + y)}{(x + y)^2} \] \[ = \frac{(x + y)(2y) - (x^2 + y^2)(1)}{(x + y)^2} \] \[ = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2} \] \[ = \frac{2xy + 2y^2 - x^2 - y^2}{(x + y)^2} \] \[ = \frac{2xy + y^2 - x^2}{(x + y)^2} \] Substitute back: \[ \frac{\partial u}{\partial y} = \frac{x + y}{x^2 + y^2} \cdot \frac{2xy + y^2 - x^2}{(x + y)^2} \] \[ = \frac{2xy + y^2 - x^2}{(x^2 + y^2)(x + y)} \] Now, we compute \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \): \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x \left( \frac{x^2 + 2xy - y^2}{(x^2 + y^2)(x + y)} \right) + y \left( \frac{2xy + y^2 - x^2}{(x^2 + y^2)(x + y)} \right) \] \[ = \frac{x(x^2 + 2xy - y^2) + y(2xy + y^2 - x^2)}{(x^2 + y^2)(x + y)} \] \[ = \frac{x^3 + 2x^2y - xy^2 + 2xy^2 + y^3 - yx^2}{(x^2 + y^2)(x + y)} \] \[ = \frac{x^3 + x^2y + xy^2 + y^3}{(x^2 + y^2)(x + y)} \] Notice that the numerator \( x^3 + x^2y + xy^2 + y^3 \) can be factored: \[ x^3 + x^2y + xy^2 + y^3 = (x + y)(x^2 + xy + y^2) \] So the expression simplifies to: \[ = \frac{(x + y)(x^2 + xy + y^2)}{(x^2 + y^2)(x + y)} \] \[ = \frac{x^2 + xy + y^2}{x^2 + y^2} \] Thus, the final answer is: \[ \boxed{\frac{x^2 + xy + y^2}{x^2 + y^2}} \]