What is the integral of (16)/(16-x^2) ?

The integral of (16)/(16-x^2) is 4(1/2 ln| x/4+1|-1/2 ln| x/4-1|)+C

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integral of (16)/(16-x^2)

\int \frac{16}{16 - x ^{2}} d x

4 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1) + C

Solution steps

\int \frac{16}{16 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int \frac{1}{16 - x ^{2}} d x

Apply Integral Substitution

= 16 \cdot \int \frac{1}{4 ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \frac{1}{4} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 16 \cdot \frac{1}{4} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{4}

= 16 \cdot \frac{1}{4} (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2})

Simplify

16 \cdot \frac{1}{4} (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2}) : 4 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1)

= 4 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1)

Add a constant to the solution

= 4 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1) + C

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What is the integral of (16)/(16-x^2) ?
The integral of (16)/(16-x^2) is 4(1/2 ln| x/4+1|-1/2 ln| x/4-1|)+C