Question Statement

Evaluate the integral:

$\int \frac{2x+1}{x^2+3}dx$

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Background and Explanation

This problem involves integrating a rational function where the numerator is not exactly the derivative of the denominator. The key strategy is to split the fraction into two parts: one where the numerator is the derivative of the denominator (yielding a logarithmic result), and another that fits the standard arctangent integral form.

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Solution

Let $I=\int \frac{2x+1}{x^2+3}dx$.

First, we split the integrand into two separate fractions:

$I=\int \left(\frac{2x}{x^2+3}+\frac{1}{x^2+3}\right)dx$

This separation allows us to evaluate two simpler integrals individually:

$I=\int \frac{2x}{x^2+3}dx+\int \frac{1}{x^2+3}dx$

First Integral: $\int \frac{2x}{x^2+3}dx$

Notice that the numerator $2x$ is exactly the derivative of the denominator $x^2+3$. This fits the logarithmic integration rule $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$. Since $x^2+3>0$ for all real $x$, we can write:

$\int \frac{2x}{x^2+3}dx=\ln \left(x^2+3\right)$

Second Integral: $\int \frac{1}{x^2+3}dx$

We rewrite the denominator to match the standard arctangent form $x^2+a^2$:

$\int \frac{1}{x^2+3}dx=\int \frac{1}{x^2+(\sqrt{3}{)}^2}dx$

Using the standard integral formula $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$ with $a=\sqrt{3}$:

$\int \frac{1}{x^2+(\sqrt{3}{)}^2}dx=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)$

Combining the results:

$I=\ln \left(x^2+3\right)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$

where $c$ is the constant of integration.

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Key Formulas or Methods Used

• Partial fraction decomposition (algebraic splitting): $\frac{2x+1}{x^2+3}=\frac{2x}{x^2+3}+\frac{1}{x^2+3}$
• Logarithmic integration rule: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
• Standard arctangent integral: $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$ (where $a=\sqrt{3}$ in this problem)

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Summary of Steps

1. Split the integrand into two fractions: $\frac{2x}{x^2+3}$ and $\frac{1}{x^2+3}$
2. Evaluate the first integral using the logarithmic rule, since the numerator is the derivative of the denominator: $\ln (x^2+3)$
3. Rewrite the second integral in the form $\frac{1}{x^2+a^2}$ by recognizing $3=(\sqrt{3}{)}^2$
4. Apply the arctangent formula with $a=\sqrt{3}$ to get $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)$
5. Combine results and add the constant of integration $c$