What is the integral of 1/(x^2-1/4) ?

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

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integral of 1/(x^2-1/4)

\int \frac{1}{x ^{2} - \frac{1}{4}} d x

- ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - \frac{1}{4}} d x

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

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

= 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

= 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}

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

Substitute back

u = \frac{x}{\frac{1}{2}}

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

Simplify

2 - \frac{ln \frac{x}{\frac{1}{2}} + 1}{2} - \frac{ln \frac{x}{\frac{1}{2}} - 1}{2} : - ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣

= - ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - ln ∣ 2 x + 1 ∣ + ln ∣ 2 x - 1 ∣ + C

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