# How do you find #int (x^2+2x+1)/((x+1)(x^2-2)) dx# using partial fractions?

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The numerator factors:

Cancel the common factor:

The denominator can be factored as the difference of two squares:

Write the partial fractions equation:

Write in integral form:

The integrals become natural logarithms:

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To find the integral of (\frac{x^2+2x+1}{(x+1)(x^2-2)}) using partial fractions, first factor the denominator completely: ((x+1)(x^2-2)).

The denominator can be factored as ( (x+1)(x-\sqrt{2})(x+\sqrt{2}) ).

The partial fraction decomposition will have the form:

(\frac{x^2+2x+1}{(x+1)(x^2-2)} = \frac{A}{x+1} + \frac{B}{x-\sqrt{2}} + \frac{C}{x+\sqrt{2}})

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

(x^2 + 2x + 1 = A(x^2 - 2) + B(x+1)(x+\sqrt{2}) + C(x+1)(x-\sqrt{2}))

Now, solve for A, B, and C by comparing coefficients.

- Substitute (x = -1): (A = \frac{1}{3}).
- Substitute (x = \sqrt{2}): (B = \frac{1}{3\sqrt{2}}).
- Substitute (x = -\sqrt{2}): (C = \frac{-1}{3\sqrt{2}}).

Now, rewrite the integral using the partial fraction decomposition and integrate term by term:

[\int \left( \frac{1}{3(x+1)} + \frac{1}{3\sqrt{2}(x-\sqrt{2})} - \frac{1}{3\sqrt{2}(x+\sqrt{2})} \right) , dx]

Integrate each term separately:

[= \frac{1}{3} \ln|x+1| + \frac{1}{3\sqrt{2}} \ln|x-\sqrt{2}| - \frac{1}{3\sqrt{2}} \ln|x+\sqrt{2}| + C]

Where (C) is the constant of integration.

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