How do you integrate $\frac{1}{(x + 1) (x + 2)}$ $1/ ((x + 1)(x + 2))$ using partial fractions?

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Answer 1 · 2017-03-27T10:19:09

$\int \frac{d x}{(x + 1) (x + 2)} = ln ∣ ∣ ∣ \frac{x + 1}{x + 2} ∣ ∣ ∣ + C$ $int dx/((x+1)(x+2)) = ln abs ((x+1)/(x+2)) +C$

Write the rational function as a sum of partial fractions:

$\frac{1}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$ $1/((x+1)(x+2)) = A/(x+1)+B/(x+2)$

$\frac{1}{(x + 1) (x + 2)} = \frac{A (x + 2) + B (x + 1)}{(x + 1) (x + 2)}$ $1/((x+1)(x+2)) = (A(x+2)+B(x+1))/((x+1)(x+2))$

$\frac{1}{(x + 1) (x + 2)} = \frac{A x + 2 A + B x + B}{(x + 1) (x + 2)}$ $1/((x+1)(x+2)) = (Ax+2A+Bx+B)/((x+1)(x+2))$

As the denominators are equal we can equate the numerators:

$1 = (A + B) x + 2 A + B$ $1 = (A+B)x+2A+B$

and equating the coefficients with the same degree in $x$ $x$ we have:

${(A+B=0),(2A+B = 1):}$

${(B=-A),(2A-A = 1):}$

${(A=1),(B= -1):}$

So that:

$1/((x+1)(x+2)) = 1/(x+1)-1/(x+2)$

and integrating:

$int dx/((x+1)(x+2)) = int dx/(x+1)-int dx/(x+2)$

$int dx/((x+1)(x+2)) = ln abs (x+1) -ln abs (x+2) +C$

and using the properties of logarithms:

$int dx/((x+1)(x+2)) = ln abs ((x+1)/(x+2)) +C$

How do you integrate 1/ ((x + 1)(x + 2))1(x+1)(x+2)1/ ((x + 1)(x + 2)) using partial fractions?