Question

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

Answer 1

-\sqrt{4x-x^2} + 4 \arcsin\left(\frac{x-2}{2}\right) + C

Answer 2

The integral to be evaluated is $I = \int \frac{x+2}{\sqrt{4x-x^2}} dx$.

This is an integral of the form $\int \frac{Px+Q}{\sqrt{ax^2+bx+c}} dx$. The general strategy is to express the numerator $Px+Q$ as a linear combination of the derivative of the quadratic expression in the denominator ($ax^2+bx+c$) and a constant term.

Step 1: Express the numerator in terms of the derivative of the quadratic expression. Let the quadratic expression in the denominator be $f(x) = 4x-x^2$. The derivative of $f(x)$ is $f'(x) = 4-2x$. We want to write the numerator $x+2$ in the form $A(4-2x) + B$. Comparing coefficients: For the $x$ term: $1 = -2A \implies A = -\frac{1}{2}$. For the constant term: $2 = 4A + B$. Substitute $A = -\frac{1}{2}$ into the constant term equation: $2 = 4\left(-\frac{1}{2}\right) + B$ $2 = -2 + B$ $B = 4$. So, we can write $x+2 = -\frac{1}{2}(4-2x) + 4$.

Step 2: Split the integral into two simpler integrals. Substitute the expression for $x+2$ back into the integral: $I = \int \frac{-\frac{1}{2}(4-2x) + 4}{\sqrt{4x-x^2}} dx$ $I = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx + 4 \int \frac{1}{\sqrt{4x-x^2}} dx$ Let $I_1 = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx$ and $I_2 = 4 \int \frac{1}{\sqrt{4x-x^2}} dx$.

Step 3: Evaluate the first integral $I_1$. For $I_1 = -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx$, let $u = 4x-x^2$. Then, $du = (4-2x)dx$. The integral becomes: $I_1 = -\frac{1}{2} \int \frac{du}{\sqrt{u}} = -\frac{1}{2} \int u^{-1/2} du$ Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$: $I_1 = -\frac{1}{2} \left( \frac{u^{-1/2+1}}{-1/2+1} \right) + C_1$ $I_1 = -\frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) + C_1$ $I_1 = -\frac{1}{2} (2\sqrt{u}) + C_1 = -\sqrt{u} + C_1$ Substitute back $u = 4x-x^2$: $I_1 = -\sqrt{4x-x^2} + C_1$.

Step 4: Evaluate the second integral $I_2$. For $I_2 = 4 \int \frac{1}{\sqrt{4x-x^2}} dx$, we need to complete the square for the quadratic expression $4x-x^2$ in the denominator. $4x-x^2 = -(x^2-4x)$ To complete the square for $x^2-4x$, we add and subtract $(4/2)^2 = 4$: $x^2-4x = (x^2-4x+4) - 4 = (x-2)^2 - 4$. So, $4x-x^2 = -((x-2)^2 - 4) = 4 - (x-2)^2$. Now substitute this into $I_2$: $I_2 = 4 \int \frac{1}{\sqrt{4-(x-2)^2}} dx$ This integral is of the standard form $\int \frac{1}{\sqrt{a^2-y^2}} dy = \arcsin\left(\frac{y}{a}\right) + C$. Here, $a^2=4 \implies a=2$, and $y=(x-2)$. So, $I_2 = 4 \arcsin\left(\frac{x-2}{2}\right) + C_2$.

Step 5: Combine the results of $I_1$ and $I_2$. $I = I_1 + I_2 = -\sqrt{4x-x^2} + 4 \arcsin\left(\frac{x-2}{2}\right) + C$, where $C = C_1+C_2$.

The final answer is $-\sqrt{4x-x^2} + 4 \arcsin\left(\frac{x-2}{2}\right) + C$.

Explanation of the solution: The integral $\int \frac{x+2}{\sqrt{4x-x^2}} dx$ is solved by first decomposing the numerator $x+2$ into two parts: one part proportional to the derivative of $4x-x^2$ (which is $4-2x$), and a constant part. This decomposition is $x+2 = -\frac{1}{2}(4-2x) + 4$. This allows the original integral to be split into two simpler integrals. The first integral, $\int \frac{-\frac{1}{2}(4-2x)}{\sqrt{4x-x^2}} dx$, is solved using a simple substitution $u = 4x-x^2$, yielding $-\sqrt{4x-x^2}$. The second integral, $\int \frac{4}{\sqrt{4x-x^2}} dx$, is solved by completing the square in the denominator, transforming $4x-x^2$ into $4-(x-2)^2$. This allows the integral to be recognized as a standard inverse sine form, resulting in $4 \arcsin\left(\frac{x-2}{2}\right)$. Combining these two results gives the final answer.

Answer: The integral of $\frac{x+2}{\sqrt{4x-x^2}}$ is $-\sqrt{4x-x^2} + 4 \arcsin\left(\frac{x-2}{2}\right) + C$.