How do you integrate $\frac{4}{(x + 2) (x + 3)}$ $4 /((x + 2)(x + 3))$ ?

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How do you integrate $\frac{4}{(x + 2) (x + 3)}$ $4 /((x + 2)(x + 3))$ ?

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Answer 1 · 2018-03-20T06:52:57

The answer is $= 4 (ln (| x + 2 |) - ln (| x + 3 |)) + C$ $=4(ln(|x+2|)-ln(|x+3|))+C$

Explanation:

Perform the decomposition into partial fractions

$\frac{4}{(x + 2) (x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3} = \frac{A (x + 3) + B (x + 2)}{(x + 2) (x + 3)}$ $(4)/((x+2)(x+3))=A/(x+2)+B/(x+3)=(A(x+3)+B(x+2))/((x+2)(x+3))$

The denominators are the same, compare the numerators

$4 = A (x + 3) + B (x + 2)$ $4=A(x+3)+B(x+2)$

Let $x = - 2$ $x=-2$ , $\Rightarrow$ $=>$ , $4 = A$ $4=A$

Let $x = - 3$ $x=-3$ , $\Rightarrow$ $=>$ , $4 = - B$ $4=-B$

Therefore,

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

$\int \frac{4 d x}{(x + 2) (x + 3)} = \int \frac{4 d x}{x + 2} + \int \frac{- 4 d x}{x + 3}$ $int(4dx)/((x+2)(x+3))=int(4dx)/(x+2)+int(-4dx)/(x+3)$

$= 4 ln (| x + 2 |) - 4 ln (| x + 3 |) + C$ $=4ln(|x+2|)-4ln(|x+3|)+C$

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How do you integrate $\frac{4}{(x + 2) (x + 3)}$ $4 /((x + 2)(x + 3))$ ?

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