Question

Use partial fractions to find the indefinite integral. $$ \int \frac{4}{x^{2}-4} d x $$

Short answer

The indefinite integral of the function is \[\ln |x-2| - \ln |x+2| + C\]

Step-by-step solution

Factoring the Denominator.

Factorize the denominator in the given integral. The factor of \(x^{2}-4\) is \((x-2)(x+2)\), since it is the difference of two squares.

Expressing the Fraction as a Partial Fraction.

A fraction with a quadratic term in the denominator can be expressed as the sum of two simpler fractions, or partial fractions. The actual partial fraction decomposition of \(\frac{4}{x^{2}-4}\) can be written as follows: \(\frac{4}{x^{2}-4}=\frac{A}{x-2}+\frac{B}{x+2}\),where A and B are constants.

Solving for the Constants A and B.

To solve for constants A and B, we make the right side identical to the left side by canceling out the denominator so that we can solve it for different values of x. We get: \(4=A(x+2)+B(x-2)\).Choosing convenient values for x, say x=2 and x=-2, will give A and B directly.Solving for A when x=2 we get:\(4=4A \Rightarrow A=1\).Next, we solve for B when x=-2:\(4=-4B \Rightarrow B=-1\).

Substitution and Integration.

Now substituting A and B into the partial fractions, we can integrate:\(\int \frac{4}{x^{2}-4} d x =\int \frac{1}{x-2} d x-\int \frac{1}{x+2} d x\),The integral of \(\frac{1}{x-a}\) is \(\ln |x-a|\). Hence, the result of the above integration is:\[\ln |x-2| - \ln |x+2| + C\], where C is the constant of integration.