- \(\displaystyle \int \frac {x+1}{\sqrt {4-x^2}} ~dx = \color {blue} {-\sqrt {4-x^2}+\sin ^{-1} (\frac {x}{2})}\)
Solution: \(\frac {x+1}{\sqrt {4-x^2}} ~dx = - \frac {1}{2} \frac {-2x}{\sqrt {4-x^2}} ~dx + \frac {1}{\sqrt {4-x^2}} ~dx\)
Solution: \(\sin (u) = \frac {1}{2} x\), \(\cos (u) ~du = \frac {1}{2} ~dx\)
Solution: \(\frac {1}{\sqrt {4-x^2}} ~dx = \frac {1}{2 \cos (u)} 2 \cdot \cos (u) ~du = du\)
Used 2019
Solution
Take \(|x| < 2\).
1. Split the integral into two parts: \[\int \frac{x+1}{\sqrt{4-x^2}} \, dx = \int \frac{x}{\sqrt{4-x^2}} \, dx + \int \frac{1}{\sqrt{4-x^2}} \, dx.\]
The integrand satisfies \(\frac{x+1}{\sqrt{4-x^2}} = \frac{x}{\sqrt{4-x^2}} + \frac{1}{\sqrt{4-x^2}}\), and the integral of a sum is the sum of the integrals.
2. \(\int \frac{x}{\sqrt{4-x^2}} \, dx = -\sqrt{4-x^2} + C_1\).
2.1. Substitute \(u = 4 - x^2 \implies du = -2x \, dx \implies x \, dx = -\frac{1}{2} \, du\).
2.2. \(\int \frac{x}{\sqrt{4-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} (2 u^{1/2}) + C_1 = -\sqrt{4-x^2} + C_1\).
3. \(\int \frac{1}{\sqrt{4-x^2}} \, dx = \arcsin\left(\frac{x}{2}\right) + C_2\).
3.1. Substitute \(x = 2\sin(\theta)\) for \(\theta \in (-\pi/2, \pi/2)\), so \(dx = 2\cos(\theta) \, d\theta\) and \(\sqrt{4-x^2} = \sqrt{4 - 4\sin^2(\theta)} = 2\cos(\theta)\).
3.2. \(\int \frac{1}{\sqrt{4-x^2}} \, dx = \int \frac{2\cos(\theta)}{2\cos(\theta)} \, d\theta = \int 1 \, d\theta = \theta + C_2 = \arcsin\left(\frac{x}{2}\right) + C_2\).
4. Combining results: \[\int \frac{x+1}{\sqrt{4-x^2}} \, dx = -\sqrt{4-x^2} + \arcsin\left(\frac{x}{2}\right) + C.\]
Adding the two antiderivatives from 2 and 3, and absorbing the constants \(C_1 + C_2\) into a single constant \(C\), gives the stated result.