int(2x-3)/(9+x^(2))dx

Identify the form: Identify the form of the integral. The integral is of the form ∫(2x-3)/(9+x^2)dx, which suggests a possible substitution for the denominator or a partial fraction decomposition. However, since the denominator is a sum of squares, we can't factor it over the reals, so partial fraction decomposition is not applicable. Instead, we will use substitution.Use substitution: Use substitution to simplify the integral. Let u = 9 + x^2. Then, du = 2x dx. Notice that the numerator, 2x-3, is not exactly du, but it contains the term 2x which is part of du. We can split the integral into two parts to handle this.Split the integral: Split the integral into two parts. ∫(2x-3)/(9+x^2)dx = ∫2x/(9+x^2)dx - ∫3/(9+x^2)dxIntegrate first part: Integrate the first part using substitution. For the first part, ∫2x/(9+x^2)dx, we can use the substitution from Step 2. u = 9 + x^2, du = 2x dx ∫2x/(9+x^2)dx = ∫1/u du The integral of 1/u with respect to u is ln|u|, so we have: ∫1/u du = ln|u| + C1 = ln|9 + x^2| + C1Integrate second part: Integrate the second part using a trigonometric substitution. For the second part, ∫3/(9+x^2)dx, we can use a trigonometric substitution. Let x = 3tan(θ), then dx = 3sec^2(θ)dθ and 9+x^2 becomes 9+9tan^2(θ) which simplifies to 9sec^2(θ). ∫3/(9+x^2)dx = ∫3/(9sec^2(θ)) * 3sec^2(θ)dθ Simplify the integrand: ∫3/(9sec^2(θ)) * 3sec^2(θ)dθ = ∫(1/3)dθ The integral of a constant is just the constant times the variable, so we have: ∫(1/3)dθ = (1/3)θ + C2Convert back to x: Convert the trigonometric result back to x. We need to express θ in terms of x. Since x = 3tan(θ), we have tan(θ) = x/3. To find θ, we take the arctan of both sides: θ = arctan(x/3) So the integral becomes: (1/3)θ + C2 = (1/3)arctan(x/3) + C2Combine the results: Combine the results from the two parts. Combining the results from Step 4 and Step 6, we get the final answer for the integral: ∫(2x-3)/(9+x^2)dx = ln|9 + x^2| + (1/3)arctan(x/3) + C where C is the constant of integration combining C1 and C2.

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Evaluate definite integrals using the chain rule

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\int\frac{2x-3}{9+x^{2}}\,dx

Identify the form: Identify the form of the integral. $\newline$ The integral is of the form $\int(2x-3)/(9+x^2)\,dx$ , which suggests a possible substitution for the denominator or a partial fraction decomposition. However, since the denominator is a sum of squares, we can't factor it over the reals, so partial fraction decomposition is not applicable. Instead, we will use substitution.
Use substitution: Use substitution to simplify the integral. $\newline$ Let $u = 9 + x^2$ . Then, $du = 2x dx$ . Notice that the numerator, $2x-3$ , is not exactly $du$ , but it contains the term $2x$ which is part of $du$ . We can split the integral into two parts to handle this.
Split the integral: Split the integral into two parts. $\newline$ \int\frac{$2$x$-3$}{$9$+x^$2$}\,dx = \int\frac{$2$x}{$9$+x^$2$}\,dx - \int\frac{$3$}{$9$+x^$2$}\,dx
Integrate first part: Integrate the first part using substitution.\(\newlineFor the first part, $\int \frac{2x}{9+x^2}\,dx$ , we can use the substitution from Step $2$ . $\newline$ $u = 9 + x^2$ , $du = 2x\,dx$ $\newline$ $\int \frac{2x}{9+x^2}\,dx = \int \frac{1}{u}\,du$ $\newline$ The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|$ , so we have: $\newline$ $\int \frac{1}{u}\,du = \ln|u| + C_1 = \ln|9 + x^2| + C_1$
Integrate second part: Integrate the second part using a trigonometric substitution. $\newline$ For the second part, $\int \frac{3}{9+x^2}\,dx$ , we can use a trigonometric substitution. Let $x = 3\tan(\theta)$ , then $dx = 3\sec^2(\theta)\,d\theta$ and $9+x^2$ becomes $9+9\tan^2(\theta)$ which simplifies to $9\sec^2(\theta)$ . $\newline$ $\int \frac{3}{9+x^2}\,dx = \int \frac{3}{9\sec^2(\theta)} \cdot 3\sec^2(\theta)\,d\theta$ $\newline$ Simplify the integrand: $\newline$ $\int \frac{3}{9\sec^2(\theta)} \cdot 3\sec^2(\theta)\,d\theta = \int(\frac{1}{3})\,d\theta$ $\newline$ The integral of a constant is just the constant times the variable, so we have: $\newline$ $\int(\frac{1}{3})\,d\theta = (\frac{1}{3})\theta + C_2$
Convert back to x: Convert the trigonometric result back to x. $\newline$ We need to express $\theta$ in terms of $x$ . Since $x = 3\tan(\theta)$ , we have $\tan(\theta) = \frac{x}{3}$ . To find $\theta$ , we take the arctan of both sides: $\newline$ $\theta = \arctan\left(\frac{x}{3}\right)$ $\newline$ So the integral becomes: $\newline$ $\frac{1}{3}\theta + C_2 = \frac{1}{3}\arctan\left(\frac{x}{3}\right) + C_2$
Combine the results: Combine the results from the two parts. $\newline$ Combining the results from Step $4$ and Step $6$ , we get the final answer for the integral: $\newline$ $\int\frac{2x-3}{9+x^2}\,dx = \ln|9 + x^2| + \left(\frac{1}{3}\right)\arctan\left(\frac{x}{3}\right) + C$ $\newline$ where $C$ is the constant of integration combining $C_1$ and $C_2$ .