int(x)/(sqrt(9-4x^(2)))dx

Identify Integral: Identify the integral to be solved. We need to find the indefinite integral of the function I = ∫(x)/(sqrt(9-4x^2)) dx.Trigonometric Substitution: Use a trigonometric substitution to simplify the integral. Let x = (3/2)sin(θ), which implies dx = (3/2)cos(θ) dθ. Then, 9 - 4x^2 = 9 - 4(3/2)^2sin^2(θ) = 9 - 9sin^2(θ) = 9cos^2(θ).Substitute Trigonometric Expressions: Substitute x and dx in the integral with the trigonometric expressions. I = ∫(3/2)sin(θ) * (3/2)cos(θ) dθ / sqrt(9cos^2(θ))Simplify Integral: Simplify the integral. I = ∫(3/2)^2sin(θ)cos(θ) dθ / (3|cos(θ)|) Since cos(θ) is always positive for θ in the range (-π/2, π/2), which corresponds to the range of arcsin(x/(3/2)), we can remove the absolute value. I = ∫(9/4)sin(θ)cos(θ) dθ / (3cos(θ)) I = (3/4)∫sin(θ) dθIntegrate with Respect: Integrate with respect to θ. I = (3/4)(-cos(θ)) + C, where C is the constant of integration.Substitute Back Variable: Substitute back the original variable x. Since x = (3/2)sin(θ), we have sin(θ) = x/(3/2) = (2/3)x. To find cos(θ), we use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1. cos(θ) = sqrt(1 - sin^2(θ)) = sqrt(1 - (2/3)^2x^2) = sqrt(1 - (4/9)x^2) = sqrt((9 - 4x^2)/9) = sqrt(9 - 4x^2)/3.Replace Cosine in Integral: Replace cos(θ) in the integral result. I = (3/4)(-sqrt(9 - 4x^2)/3) + C I = -1/4 * sqrt(9 - 4x^2) + C

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$\int \frac{x}{\sqrt{9-4 x^{2}}} d x$

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Find indefinite integrals using the substitution and by parts

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\int \frac{x}{\sqrt{9-4 x^{2}}} d x

Identify Integral: Identify the integral to be solved. $\newline$ We need to find the indefinite integral of the function $I = \int\frac{x}{\sqrt{9-4x^2}} \, dx$ .
Trigonometric Substitution: Use a trigonometric substitution to simplify the integral. $\newline$ Let $x = \frac{3}{2}\sin(\theta)$ , which implies $dx = \frac{3}{2}\cos(\theta) d\theta$ . Then, $9 - 4x^2 = 9 - 4\left(\frac{3}{2}\right)^2\sin^2(\theta) = 9 - 9\sin^2(\theta) = 9\cos^2(\theta)$ .
Substitute Trigonometric Expressions: Substitute $x$ and $dx$ in the integral with the trigonometric expressions. $I = \int\left(\frac{3}{2}\right)\sin(\theta) \cdot \left(\frac{3}{2}\right)\cos(\theta) d\theta / \sqrt{9\cos^2(\theta)}$
Simplify Integral: Simplify the integral. $\newline$ $I = \int\left(\frac{3}{2}\right)^2\sin(\theta)\cos(\theta) d\theta / \left(3|\cos(\theta)|\right)$ $\newline$ Since $\cos(\theta)$ is always positive for $\theta$ in the range $(-\pi/2, \pi/2)$ , which corresponds to the range of $\arcsin\left(\frac{x}{3/2}\right)$ , we can remove the absolute value. $\newline$ $I = \int\left(\frac{9}{4}\right)\sin(\theta)\cos(\theta) d\theta / \left(3\cos(\theta)\right)$ $\newline$ $I = \left(\frac{3}{4}\right)\int\sin(\theta) d\theta$
Integrate with Respect: Integrate with respect to $\theta$ . $I = \left(\frac{3}{4}\right)(-\cos(\theta)) + C$ , where $C$ is the constant of integration.
Substitute Back Variable: Substitute back the original variable $x$ . $\newline$ Since $x = \frac{3}{2}\sin(\theta)$ , we have $\sin(\theta) = \frac{x}{\frac{3}{2}} = \frac{2}{3}x$ . To find $\cos(\theta)$ , we use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ . $\newline$ $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{2}{3}\right)^2x^2} = \sqrt{1 - \frac{4}{9}x^2} = \sqrt{\frac{9 - 4x^2}{9}} = \frac{\sqrt{9 - 4x^2}}{3}$ .
Replace Cosine in Integral: Replace $\cos(\theta)$ in the integral result. $I = \left(\frac{3}{4}\right)\left(-\sqrt{9 - 4x^2}/3\right) + C$ $I = -\frac{1}{4} \sqrt{9 - 4x^2} + C$