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

The integral of 1/(x^2+1/4) is 2arctan(2x)+C

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

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

2 arctan (2 x) + 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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 arctan (u)

Substitute back

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

= 2 arctan (\frac{x}{\frac{1}{2}})

Simplify

= 2 arctan (2 x)

Add a constant to the solution

= 2 arctan (2 x) + C

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