How do you integrate $\int \frac{1}{x^{2} - x - 20} d x$ $int 1/(x^2-x-20) dx$ using partial fractions?

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How do you integrate $\int \frac{1}{x^{2} - x - 20} d x$ $int 1/(x^2-x-20) dx$ using partial fractions?

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Answer 1 · 2018-05-07T02:32:52

$\frac{1}{9} ln | x - 5 | - \frac{1}{9} ln | x + 4 | + C$ $1/9ln|x-5|-1/9ln|x+4|+C$

Explanation:

Factor the denominator of the integrand:

$x^{2} - x - 20 = (x - 5) (x + 4)$ $x^2-x-20=(x-5)(x+4)$

Use partial fraction decomposition on the simplified integrand:

$\frac{1}{(x - 5) (x + 4)} = \frac{A}{x - 5} + \frac{B}{x + 4}$ $1/((x-5)(x+4))=A/(x-5)+B/(x+4)$

Add up the right side:

$\frac{1}{(x - 5) (x + 4)} = \frac{A (x + 4) + B (x - 5)}{(x - 5) (x + 4)}$ $1/((x-5)(x+4))=(A(x+4)+B(x-5))/((x-5)(x+4))$

Equate numerators:

$1 = A (x + 4) + B (x - 5)$ $1=A(x+4)+B(x-5)$

We need to find $A, B .$ $A,B.$ We can do this by plugging in values of $x$ $x$ that send one term to $0$ $0$ and keep the other:

$x = - 4 :$ $x=-4:$

$1 = - 9 B, B = - \frac{1}{9}$ $1=-9B, B=-1/9$

$x = 5 :$ $x=5:$

$1 = 9 A, A = \frac{1}{9}$ $1=9A, A=1/9$

Thus, our integral becomes

$\int \frac{\frac{1}{9}}{x - 5} - \frac{\frac{1}{9}}{x + 4} d x = \frac{1}{9} ln | x - 5 | - \frac{1}{9} ln | x + 4 | + C$ $int(1/9)/(x-5)-(1/9)/(x+4)dx=1/9ln|x-5|-1/9ln|x+4|+C$

Answer 2 · 2018-05-07T02:36:23

$\frac{1}{9} \cdot ln | x - 5 | - \frac{1}{9} \cdot ln | x + 4 |$ $1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)$

Explanation:

$\frac{1}{(x - 5) (x + 4)} = \frac{1}{9 \cdot (x - 5)} - \frac{1}{9 \cdot (x + 4)}$ $1/((x-5)(x+4)) = 1/(9*(x-5)) - 1/(9*(x+4))$

$\int \frac{1}{(x - 5) (x + 4)} = \int \frac{1}{9 \cdot (x - 5)} - \frac{1}{9 \cdot (x + 4)} d x$ $int1/((x-5)(x+4)) = int1/(9*(x-5)) - 1/(9*(x+4))dx$

$\int \frac{1}{(x - 5) (x + 4)} = \frac{1}{9} \cdot ln | x - 5 | - \frac{1}{9} \cdot ln | x + 4 |$ $int1/((x-5)(x+4)) = 1/9 * lnabs(x-5)- 1/9 * lnabs(x+4)$

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How do you integrate $\int \frac{1}{x^{2} - x - 20} d x$ $int 1/(x^2-x-20) dx$ using partial fractions?

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