What is the integral of 1/(2x^2-16) ?

The integral of 1/(2x^2-16) is -1/(8sqrt(2))(ln| x/(2sqrt(2))+1|-ln| x/(2sqrt(2))-1|)+C

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integral of 1/(2x^2-16)

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

- \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

= \frac{1}{4 2} (- \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}

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

Substitute back

u = \frac{2}{4} x

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

Simplify

\frac{1}{4 2} - \frac{ln \frac{2}{4} x + 1}{2} - \frac{ln \frac{2}{4} x - 1}{2} : - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

= - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

Add a constant to the solution

= - \frac{1}{8 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

\int \frac{4 - x ^{2}}{x} d x

x \to - 2 lim (\frac{x}{( x + 2 ) ^{2}})

x \to - 4 lim ((x)^{2})

\int \frac{1}{x ^{2} + 3} d x

θ \to 0 lim (θ cos (\frac{3}{θ}))

What is the integral of 1/(2x^2-16) ?
The integral of 1/(2x^2-16) is -1/(8sqrt(2))(ln| x/(2sqrt(2))+1|-ln| x/(2sqrt(2))-1|)+C