What is the integral of 1/((x+1/2)^2+3/4) ?

The integral of 1/((x+1/2)^2+3/4) is 2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+C

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integral of 1/((x+1/2)^2+3/4)

\int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

\frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

Solution steps

\int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

(x + \frac{1}{2})^{2} + \frac{3}{4} = \frac{3}{4} (\frac{( x + \frac{1}{2} ) ^{2}}{\frac{3}{4}} + 1)

= \int \frac{1}{\frac{3}{4} ( \frac{( x + \frac{1}{2} ) ^{2}}{\frac{3}{4}} + 1 )} d x

Take the constant out:

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

= \frac{1}{\frac{3}{4}} \cdot \int \frac{1}{\frac{( x + \frac{1}{2} ) ^{2}}{\frac{3}{4}} + 1} d x

Simplify

= \frac{4}{3} \cdot \int \frac{1}{\frac{( x + \frac{1}{2} ) ^{2}}{\frac{3}{4}} + 1} d x

Apply u-substitution

= \frac{4}{3} \cdot \int \frac{3}{4 u ^{2} + 3} d u

Take the constant out:

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

= \frac{4}{3} \cdot 3 \cdot \int \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= \frac{4}{3} \cdot 3 \cdot \int \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{4}{3} \cdot 3 \cdot \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{4}{3} \cdot 3 \cdot \frac{1}{2 3} arctan (v)

Substitute back

= \frac{4}{3} \cdot 3 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2}))

Simplify

\frac{4}{3} \cdot 3 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2})) : \frac{2}{3} arctan (\frac{2 x + 1}{3})

= \frac{2}{3} arctan (\frac{2 x + 1}{3})

Add a constant to the solution

= \frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

What is the integral of 1/((x+1/2)^2+3/4) ?
The integral of 1/((x+1/2)^2+3/4) is 2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+C