Evaluate the integral and express your answer in simplest form. int(3)/(1+9x^(2))dx Answer:

Given Integral: We are given the integral to evaluate: ∫(3)/(1+9x^(2))dx To solve this integral, we can use a trigonometric substitution because the denominator has the form of 1 + a^2x^2, which suggests using the substitution x = (1/a)tan(θ). Let's choose x = (1/3)tan(θ), so that 9x^2 becomes tan^2(θ).Trigonometric Substitution: First, we need to find dx in terms of dθ. Differentiating x = (1/3)tan(θ) with respect to θ gives us: dx/dθ = (1/3)sec^2(θ) dx = (1/3)sec^2(θ)dθFinding dx in terms of dθ: Now we substitute x with (1/3)tan(θ) and dx with (1/3)sec^2(θ)dθ in the integral: ∫(3)/(1+9x^(2))dx = ∫(3)/(1+tan^2(θ)) * (1/3)sec^2(θ)dθ Simplify the integral using the trigonometric identity 1 + tan^2(θ) = sec^2(θ): ∫(3)/(sec^2(θ)) * (1/3)sec^2(θ)dθSubstituting x and dx in the integral: The sec^2(θ) terms cancel out, leaving us with: ∫dθ This integral is straightforward to evaluate: ∫dθ = θ + CSimplifying the integral: Now we need to express θ back in terms of x. From our substitution x = (1/3)tan(θ), we can write: θ = arctan(3x) So the integral becomes: θ + C = arctan(3x) + CEvaluating the integral: We have now found the antiderivative of the given function. The final answer in simplest form is: arctan(3x) + C

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Evaluate the integral and express your answer in simplest form.

int(3)/(1+9x^(2))dx
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Evaluate the integral and express your answer in simplest form. $\newline$ $\int \frac{3}{1+9 x^{2}} d x$ $\newline$ Answer:

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Q. Evaluate the integral and express your answer in simplest form.

\newline

\int \frac{3}{1+9 x^{2}} d x

\newline

Answer:

Given Integral: We are given the integral to evaluate: $\newline$ $\int\frac{3}{1+9x^{2}}dx$ $\newline$ To solve this integral, we can use a trigonometric substitution because the denominator has the form of $1 + a^2x^2$ , which suggests using the substitution $x = \frac{1}{a}\tan(\theta)$ . $\newline$ Let's choose $x = \frac{1}{3}\tan(\theta)$ , so that $9x^2$ becomes $\tan^2(\theta)$ .
Trigonometric Substitution: First, we need to find $dx$ in terms of $d\theta$ . Differentiating $x = \frac{1}{3}\tan(\theta)$ with respect to $\theta$ gives us: $\newline$ $\frac{dx}{d\theta} = \frac{1}{3}\sec^2(\theta)$ $\newline$ $dx = \frac{1}{3}\sec^2(\theta)d\theta$
Finding $dx$ in terms of $d\theta:$ Now we substitute $x$ with $(1/3)\tan(\theta)$ and $dx$ with $(1/3)\sec^2(\theta)d\theta$ in the integral: $\newline$ $\int\frac{3}{1+9x^{2}}dx = \int\frac{3}{1+\tan^2(\theta)} \cdot \frac{1}{3}\sec^2(\theta)d\theta$ $\newline$ Simplify the integral using the trigonometric identity $1 + \tan^2(\theta) = \sec^2(\theta):$ $\newline$ $\int\frac{3}{\sec^2(\theta)} \cdot \frac{1}{3}\sec^2(\theta)d\theta$
Substituting $x$ and $dx$ in the integral: The $\sec^2(\theta)$ terms cancel out, leaving us with: $\int d\theta$ This integral is straightforward to evaluate: $\int d\theta = \theta + C$
Simplifying the integral: Now we need to express $\theta$ back in terms of $x$ . From our substitution $x = \frac{1}{3}\tan(\theta)$ , we can write: $\theta = \arctan(3x)$ So the integral becomes: $\theta + C = \arctan(3x) + C$
Evaluating the integral: We have now found the antiderivative of the given function. The final answer in simplest form is: $\arctan(3x) + C$