How do you find the antiderivative of #int 1/(4-x^2)dx#?

Question

Ezekiel Alexander · Accepted Answer

#1/4ln|(x+2)/(x-2)|+C#.

#I=int1/(4-x^2)dx=int1/{(2+x)(2-x)dx# #=1/4int4/{(2+x)(2-x)dx# #=1/4int{(2+x)+(2-x)}/((2+x)(2-x))dx# #=1/4int[cancel((2+x))/{cancel((2+x))(2-x)}+cancel((2-x))/{cancel((2-x))(2+x)}]# #=1/4int(-1/(x-2)+1/(x+2))dx# Knowing that, for #a!=0, int1/(ax+b)dx=1/aln|ax+b|+c,# #I=1/4(-ln|x-2|+ln|x+2|)# #=1/4ln|(x+2)/(x-2)|+C#.

Cameron Alexander · Answer

undefined To find the antiderivative of $ \int \frac{1}{4-x^2} \, dx $, you can use the technique of partial fraction decomposition. First, factor the denominator as $ (2+x)(2-x) $. Then, express $ \frac{1}{4-x^2} $ as the sum of two fractions with undetermined constants:

$$ \frac{1}{4-x^2} = \frac{A}{2+x} + \frac{B}{2-x} $$

Solve for $ A $ and $ B $ by equating numerators:

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

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

Equating coefficients gives:

$$ 2A + 2B = 1 $$
$$ -A + B = 0 $$

Solving this system of equations yields $ A = \frac{1}{4} $ and $ B = \frac{1}{4} $.

Thus, we can rewrite the integral as:

$$ \int \frac{1}{4-x^2} \, dx = \int \left( \frac{\frac{1}{4}}{2+x} + \frac{\frac{1}{4}}{2-x} \right) \, dx $$

Now integrate each term separately:

$$ \int \frac{1}{4-x^2} \, dx = \frac{1}{4} \ln|2+x| - \frac{1}{4} \ln|2-x| + C $$

Where $ C $ is the constant of integration.