How do you find $\int \frac{x}{(x + 2) (x + 9)} d x$ $int x/((x+2)(x+9))dx$ using partial fractions?

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How do you find $\int \frac{x}{(x + 2) (x + 9)} d x$ $int x/((x+2)(x+9))dx$ using partial fractions?

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Answer 1 · 2015-11-03T02:31:03

$\frac{9}{7} ln (x + 9) - \frac{2}{7} ln (x + 2) +$ $9/7ln(x+9) - 2/7ln(x+2)+$ constant

Explanation:

The equation to be integrated can be written as:

$\frac{x}{(x + 2) (x + 9)} = \frac{A}{x + 2} + \frac{B}{x + 9} = \frac{A (x + 9) + B (x + 2)}{(x + 2) (x + 9)}$ $x/((x+2)(x+9))=A/(x+2)+B/(x+9)=(A(x+9)+B(x+2))/((x+2)(x+9)$

Equating coefficients

$x \to A (x + 9) + B (x + 2) = (A + B) x + (9 A + 2 B)$ $x->A(x+9)+B(x+2)=(A+B)x+(9A+2B)$

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

Therefore, dividing by $x$ $x$ :

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

Additionally,

$9 A + 2 B = 0$ $9A+2B=0$ $(2)$ $(2)$

Simultaneous Equations

$(2) - 2 \times (1)$ $(2)-2times(1)$

$9 A + 2 B = 0$ $9A+2B=0$ $(2)$ $(2)$
$- 2 A - 2 B = - 2$ $-2A-2B=-2$ $(1)$ $(1)$

Cancel out the $B$ $B$ terms:

$7 A = - 2$ $7A=-2$

$A = - \frac{2}{7}$ $A=-2/7$

Subbing $A$ $A$ back into $(1)$ $(1)$ :

$B = \frac{9}{7}$ $B=9/7$

Integrating

Subbing $A$ $A$ and $B$ $B$ back into the original equation:

$\int \frac{9}{7 (x + 9)} - \frac{2}{7 (x + 2)} d x$ $int9/(7(x+9))-2/(7(x+2))dx$

$= \int \frac{9}{7 (x + 9)} d x - \int \frac{2}{7 (x + 2)} d x$ $=int9/(7(x+9))dx-int2/(7(x+2))dx$

$= \frac{9}{7} \int \frac{1}{x + 9} d x - \frac{2}{7} \int \frac{1}{x + 2} d x$ $=9/7int1/(x+9)dx-2/7int1/(x+2)dx$

$= \frac{9}{7} ln (x + 9) +$ $=9/7ln(x+9) +$ constant $- \frac{2}{7} ln (x + 2) +$ $-2/7ln(x+2)+$ constant

$= \frac{9}{7} ln (x + 9) - \frac{2}{7} ln (x + 2) +$ $=9/7ln(x+9) - 2/7ln(x+2)+$ constant

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How do you find $\int \frac{x}{(x + 2) (x + 9)} d x$ $int x/((x+2)(x+9))dx$ using partial fractions?

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