How do you integrate $\int \frac{3 x + 2}{(x - 1) (x + 4)}$ $int (3x + 2) / [(x - 1)(x + 4)]$ using partial fractions?

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How do you integrate $\int \frac{3 x + 2}{(x - 1) (x + 4)}$ $int (3x + 2) / [(x - 1)(x + 4)]$ using partial fractions?

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Answer 1 · 2016-02-21T19:24:02

We have that

$\frac{3 x + 2}{(x - 1) (x + 4)} = \frac{2}{x + 4} + \frac{1}{x - 1}$ $(3x+2)/[(x-1)(x+4)]=2/(x+4)+1/(x-1)$

Hence

$\int \frac{3 x + 2}{(x - 1) (x + 4)} d x = \int \frac{2}{x + 4} d x + \int \frac{1}{x - 1} d x = 2 \cdot log | x + 4 | + log | x - 1 | + c$ $int (3x+2)/[(x-1)(x+4)]dx=int 2/(x+4)dx+int 1/(x-1)dx= 2*logabs(x+4)+logabs(x-1)+c$

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How do you integrate $\int \frac{3 x + 2}{(x - 1) (x + 4)}$ $int (3x + 2) / [(x - 1)(x + 4)]$ using partial fractions?

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How do you integrate ∫3x+2(x−1)(x+4)int (3x + 2) / [(x - 1)(x + 4)] using partial fractions?

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How do you integrate $\int \frac{3 x + 2}{(x - 1) (x + 4)}$ $int (3x + 2) / [(x - 1)(x + 4)]$ using partial fractions?