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# how to integrate (3x+2)/(x^2+6x+8)"?

Indefinate integral

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- 1 decade agoFavorite Answer
Notice that the denominator can be factored as (x + 2)(x + 4). This tells us that we should use partial fractions.

Suppose (3x + 2)/[(x + 2)(x + 4)] = A/(x + 2) + B/(x + 4). Then

3x + 2 = A(x + 4) + B(x + 2) = Ax + 4A + Bx + 2B = (A + B)x + 4A + 2B

We get that A + B = 3 and 4A + 2B = 2. If we multiply the first equation by 2 and subtract it from the second equation, we get that 2A = -4 so that A = -2. Thus -2 + B = 3 so that B = 5. Hence

∫(3x + 2)/(x^2 + 6x + 8) dx

= ∫[5/(x + 4) - 2/(x + 2)]dx

= 5ln|x + 4| - 2ln|x + 2| + C.

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