Question

integrate: x+2/sqrt (4x+x^2)

Answer 1

sqrt(x^2+4x) + C

Answer 2

To integrate the given function $\frac{x+2}{\sqrt{4x+x^2}}$, we can use a substitution method by observing the relationship between the numerator and the expression inside the square root.

Let the integral be $I = \int \frac{x+2}{\sqrt{4x+x^2}} dx$.

Consider the expression inside the square root in the denominator: $4x+x^2$. Let $u = x^2+4x$. Now, find the derivative of $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(x^2+4x) = 2x+4$. We can factor out 2 from the derivative: $du = (2x+4)dx = 2(x+2)dx$.

From this, we can express $(x+2)dx$ in terms of $du$: $(x+2)dx = \frac{1}{2}du$.

Now, substitute $u$ and $du$ into the integral: $I = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du$ $I = \frac{1}{2} \int u^{-1/2} du$.

Now, integrate $u^{-1/2}$ with respect to $u$ using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $I = \frac{1}{2} \left( \frac{u^{-1/2+1}}{-1/2+1} \right) + C$ $I = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) + C$ $I = \frac{1}{2} (2u^{1/2}) + C$ $I = u^{1/2} + C$ $I = \sqrt{u} + C$.

Finally, substitute back $u = x^2+4x$: $I = \sqrt{x^2+4x} + C$.

To verify the result, differentiate $\sqrt{x^2+4x}$: $\frac{d}{dx}(\sqrt{x^2+4x}) = \frac{1}{2\sqrt{x^2+4x}} \cdot \frac{d}{dx}(x^2+4x)$ $= \frac{1}{2\sqrt{x^2+4x}} \cdot (2x+4)$ $= \frac{2(x+2)}{2\sqrt{x^2+4x}}$ $= \frac{x+2}{\sqrt{x^2+4x}}$. This matches the original integrand, confirming the solution.

The final answer is $\boxed{\sqrt{x^2+4x} + C}$.

Explanation of the solution: The integral $\int \frac{x+2}{\sqrt{4x+x^2}} dx$ is solved by recognizing that the numerator $(x+2)$ is directly related to the derivative of the term inside the square root, $x^2+4x$. Let $u = x^2+4x$. Then $du = (2x+4)dx = 2(x+2)dx$. This means $(x+2)dx = \frac{1}{2}du$. Substituting these into the integral transforms it into $\int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{2} \int u^{-1/2} du$. Integrating $u^{-1/2}$ gives $\frac{u^{1/2}}{1/2} = 2\sqrt{u}$. So, the integral becomes $\frac{1}{2} (2\sqrt{u}) + C = \sqrt{u} + C$. Substituting back $u = x^2+4x$ yields the final result $\sqrt{x^2+4x} + C$.