Math / Calculus

Question $\int \frac{1}{16-x^{2}} d x$

Studdy Solution

STEP 1

Assumptions
1. The integral to be solved is $\int \frac{1}{16-x^{2}} \, dx$ .
2. The function is a rational function that can be integrated using a trigonometric substitution or recognizing a standard form.
3. The form of the integral resembles the derivative of the inverse hyperbolic tangent function or the inverse trigonometric function.

STEP 2

Recognize the standard form of the integral. The integral $\int \frac{1}{a^2 - x^2} \, dx$ can be solved using the formula:
$\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$
where $a^2 = 16$ in this case.

STEP 3

Identify the value of $a$ . Since $a^2 = 16$ , we have $a = 4$ .

STEP 4

Substitute $a = 4$ into the formula:
$\int \frac{1}{16 - x^2} \, dx = \frac{1}{2 \times 4} \ln \left| \frac{4 + x}{4 - x} \right| + C$

STEP 5

Simplify the expression:
$\int \frac{1}{16 - x^2} \, dx = \frac{1}{8} \ln \left| \frac{4 + x}{4 - x} \right| + C$
This is the solution to the integral.

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