# How do you integrate #int (3-2x)/(x(x²+3))# using partial fractions?

I broke down the integrand into simpler fractions.

Following denominator expansion,

Thus,

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To integrate (\frac{3-2x}{x(x^2+3)}) using partial fractions, first express the fraction as the sum of simpler fractions. Perform the decomposition as follows:

[\frac{3-2x}{x(x^2+3)} = \frac{A}{x} + \frac{Bx + C}{x^2+3}]

Solve for (A), (B), and (C) by equating numerators:

[3 - 2x = A(x^2 + 3) + (Bx + C)x]

Equating coefficients of like terms:

[3 - 2x = Ax^2 + 3A + Bx^2 + Cx]

This equation holds for all (x), so coefficients of like powers of (x) must be equal:

For the (x^2) terms: [A + B = 0]

For the (x) terms: [C = -2]

For the constant terms: [3A = 3]

Solve these equations to find (A), (B), and (C):

[A = 1] [B = -1] [C = -2]

So, the partial fraction decomposition is: [\frac{3-2x}{x(x^2+3)} = \frac{1}{x} - \frac{x - 2}{x^2+3}]

Now, integrate each term separately:

[\int \frac{1}{x} dx - \int \frac{x - 2}{x^2+3} dx]

[\ln|x| - \frac{1}{2}\ln(x^2+3) - \sqrt{3}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C]

So, the integral of (\frac{3-2x}{x(x^2+3)}) using partial fractions is:

[\ln|x| - \frac{1}{2}\ln(x^2+3) - \sqrt{3}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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