# How do you integrate #int (3x^2+x+4)/((x^2+2)(x^2+1))# using partial fractions?

The integral

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To integrate (\frac{{3x^2 + x + 4}}{{(x^2 + 2)(x^2 + 1)}}) using partial fractions, first, express the rational function as a sum of partial fractions.

[\frac{{3x^2 + x + 4}}{{(x^2 + 2)(x^2 + 1)}} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 1}]

Multiply both sides by ((x^2 + 2)(x^2 + 1)) to clear the denominators:

[3x^2 + x + 4 = A(x^2 + 1) + B(x^2 + 2)]

Expand and equate coefficients:

[3x^2 + x + 4 = Ax^2 + A + Bx^2 + 2B]

Matching coefficients:

[3 = A + B] [1 = B] [4 = A + 2B]

Solving these equations, you get (A = 1) and (B = 2).

So, (\frac{{3x^2 + x + 4}}{{(x^2 + 2)(x^2 + 1)}} = \frac{1}{{x^2 + 2}} + \frac{2}{{x^2 + 1}}).

Now, integrate each term separately:

[\int \frac{{3x^2 + x + 4}}{{(x^2 + 2)(x^2 + 1)}} dx = \int \frac{1}{{x^2 + 2}} dx + \int \frac{2}{{x^2 + 1}} dx]

[\int \frac{1}{{x^2 + 2}} dx = \frac{1}{\sqrt{2}} \arctan{\left(\frac{x}{\sqrt{2}}\right)} + C_1]

[\int \frac{2}{{x^2 + 1}} dx = 2 \arctan{x} + C_2]

Where (C_1) and (C_2) are constants of integration.

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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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