Integrate dx/(4-x^2)
Understand the Problem
The question asks to find the integral of the function 1/(4-x^2) with respect to x. This integral can be solved using partial fraction decomposition.
Answer
$\frac{1}{4} \ln \left| \frac{2+x}{2-x} \right| + C$
Answer for screen readers
$\frac{1}{4} \ln \left| \frac{2+x}{2-x} \right| + C$
Steps to Solve
- Partial Fraction Decomposition
We need to decompose the fraction $\frac{1}{4-x^2}$ into simpler fractions. Notice that $4-x^2 = (2-x)(2+x)$. Thus, we want to find constants $A$ and $B$ such that $$ \frac{1}{4-x^2} = \frac{A}{2-x} + \frac{B}{2+x} $$
- Solve for A and B
Multiplying both sides by $(2-x)(2+x)$ gives: $$ 1 = A(2+x) + B(2-x) $$ We can solve for $A$ and $B$ by substituting convenient values for $x$.
If $x = 2$, then $1 = A(2+2) + B(2-2) \Rightarrow 1 = 4A \Rightarrow A = \frac{1}{4}$.
If $x = -2$, then $1 = A(2-2) + B(2-(-2)) \Rightarrow 1 = 4B \Rightarrow B = \frac{1}{4}$.
- Rewrite the Integral
Now we can rewrite the integral as: $$ \int \frac{1}{4-x^2} , dx = \int \left( \frac{1/4}{2-x} + \frac{1/4}{2+x} \right) dx = \frac{1}{4} \int \left( \frac{1}{2-x} + \frac{1}{2+x} \right) dx $$
- Integrate
Now integrate each term separately: $$ \frac{1}{4} \int \frac{1}{2-x} , dx + \frac{1}{4} \int \frac{1}{2+x} , dx $$ For the first integral, let $u = 2-x$, then $du = -dx$. So, $\int \frac{1}{2-x} , dx = -\int \frac{1}{u} , du = -\ln|u| = -\ln|2-x|$.
For the second integral, let $v = 2+x$, then $dv = dx$. So, $\int \frac{1}{2+x} , dx = \int \frac{1}{v} , dv = \ln|v| = \ln|2+x|$.
Therefore, the integral is: $$ \frac{1}{4} \left( -\ln|2-x| + \ln|2+x| \right) + C = \frac{1}{4} \ln \left| \frac{2+x}{2-x} \right| + C $$
$\frac{1}{4} \ln \left| \frac{2+x}{2-x} \right| + C$
More Information
The integral can be expressed in different logarithmic forms, but they are equivalent. Using logarithm rules allows us to combine the two logarithms into one.
Tips
A common mistake is forgetting the absolute value when integrating $1/u$, it should be $\ln|u|$. Also, forgetting the negative sign when integrating $\frac{1}{2-x}$ is common. And of course, not including the constant of integration $C$.
Another common mistake is messing up the partial fraction decomposition, and incorrectly calculating A and B. Especially, when the roots are negative, the partial fraction decomposition is often done wrong.
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